LIBRARY OF CONGRESS, 

COPYRIGHT OFFICE. 

No registration (*m> f this book 
as a preliminary to copyright protec- 
tion has been found. 

Forwarded to Order Division _JMAR % W\Q 

(Apr. 5, 1901—5,000.) 




LUDLOW TEXTILE 
ARITHMETIC 



BY 



JOSEPH J. EATON, S. B. 
u 

Director, Ludlow Textile School, 
Ludlow, Mass. 

AND 

RICHARD BRADY 
i 

Overseer, Ludlow Manufacturing Associates, 
Ludlow, Mass. 



1908 
The C. R. Kaplinger Co. 

SPRINGFIELD, MASS. 






^ 



f^eceivsd from 

Copyright Office, 

MAR i 191U 



V 



J0-SZ1O 



Preface 



This book is arranged for the use of the Ludlow Textile 
School. 

It shows by practical illustrations and examples how the 
different arithmetical processes are applied to mill work 
in this particular branch of the textile industry. • 

The arithmetic is not intended to take the place of a com- 
plete arithmetic but at the same time a complete set of 
abstract examples are arranged according to grade and the 
concrete examples are extremely practical. 

The general necessity for an early understanding of 
decimals and proportion caused the placing of those matters 
at an early part of the book. Fractions have been treated 
in detail although they are used to a limited extent in the 
mills. 

In many cases the language used is that of the mill, 
consequently some words, usually explained in the notes, 
have other than the generally accepted meanings. 

Only the calculations in general use in the Ludlow Mills, 
which have been tested practically, are included in the 
chapter on General Mill Work. 

The explanations have been made as simple as possible 

Later, it is intended to publish a school description of 
the various processes of manufacture in vogue in the jute 
and hemp industry which will contain a number of more 
advanced problems. 

Some of the examples in this book have been taken from 
the following arithmetics, Bradbury and Eaton's, Thomp- 
son's, Smith's, and Wentworth's. Other authorities con- 
sulted were Carter, Sharp, and Leggatt. * 

As this is the first attempt to publish an arithmetic 
relating to this particular branch of the textile industry, 
any suggestion for improvement will gladly be received. 



Contents 



PAGE 

Vocabulary, 

Notation and Numeration, 1 

Addition, ■ 2 

Subtraction 8 

Multiplication, . » 9 

Division, 11 

Miscellaneous Examples 14 

Analysis, 17 

Factors and Multiples, 19 

Cancellation, . 23 

United States Money, 24 

Decimals, 26 

Proportion, 38 

Fractions, 42 

Measurements; Weighing; Ludlow Yarn Measure ; 

Tables, . 57 

Alligation, 71 

Percentage, 73 

Roots, 79 

General Mill Work; Belting; Horse Power; 

R. P. M. ; Rollers; Pulleys; Surface Speeds; 

Drafts; Twists, 82 

Short Processes, 102 

Answers 103 



Vocabulary 



Bagging. — A coarse fabric made from jute for covers 
of bales'of cotton. A product of the Ludlow Mills. 

Bale. — A large package prepared for transportation or 
storage. All the fibre arrives in Ludlow in bales. 

Baling. — An unfinished twine made in the Ludlow 
Mills and used chiefly for sewing bales. 

Baller. — -The operator of the balling machines. The 
latter are used for the purpose of winding the twine into 
balls. 

Batch. — A quantity of hemp, or jute placed together 
at one time, over each layer of which is spread a certain 
amount of liquor composed chiefly of oil. This batch is 
allowed to lay a certain length of time in order that the 
liquor may penetrate through all of the fibre. 

Breaker Card. — -This is a machine on which the fibre 
is first broken into tow. It is called breaker card to dis- 
tinguish it from a finer machine which is known as a finisher 
card. 

Card.- — A machine used to comb and cleanse the fibres 
and to form them into sliver. 

Carpet Yarn. — Used in the manufacture of carpets, 
one of the main products of the Ludlow Mills, made from 
hemp, flax, and jute fibres. 

Creel. — That part of the machine in which is placed 
the rove bobbins while they are being unwound in a spinning 
frame or other machine. 

Cut. — 300 yards of yarn. Same as lea. 

Cuts per Spindle. — The number of yards of spun yarn 
produced from each spinning spindle. 

Doff. — 1. To take the full bobbins from a machine 
and replace them with empty bobbins. 

2. The number of full bobbins on a spinning or roving 
or twisting frame. 



viii VOCABULARY 

Doffer. — 1. Girls who replace the full bobbins with 
empty ones. 

2. The roller which takes the fibre of! the card cylinder. 

Draft. — The amount that the material is drawn out 
between the receiving and delivery rollers. 

Feed Roller. — The roller which receives or "feeds" 
the material into a carding machine. 

Filling. — The weft of woven fabrics. 

Fluted Roller.- — Rollers with corrugated surfaces used 
as retaining or drawing rollers of wet spinning frames and 
on other textile machines. 

Flyer. — An inverted, two pronged fork, which, fastened 
to the top of a spindle, is used for putting the twist into 
yarn and for winding the latter upon a bobbin. 

Gill. — The brass stock into which the pins are driven 
which are used for combing the fibre on drawing and roving 

frames. 

Hackle. — Used for combing flax, hemp, jute, and other 
fibres. 

Hank. — Twelve cuts or 3600 yards of yarn. 

Hemp. — Many kinds are grown in different parts of the 
world. The tough and strong fibre obtained from these 
plants is largely used in the manufacture of yarns, twines, 
and ropes. The principal source of supply for American 
hemp is the Ohio valley. Hemp of other kinds is imported 
from Italy and from Russia. The coarser kind, sisal, is 
imported from Yucatan, Mexico. Manila hemp comes 
from the Philippines. 

Horse Power. — The power necessary to raise 33000 
pounds one foot in one minute. 

Jute. — An annual Asiatic plant of the linden family. 
The fibre obtained from the inner bark of this plant, is 
used in Ludlow for the manufacture of yarns and twines. 

Jenny. — One of the machines used in Ludlow in the 
manufacture of binder twines. 

Lap. — A number of slivers from a breaker card wound 
around a stick made in the form of an axle. This is done 
for convenience in handling and weighing. 

Lbs. per Spyndle. — The number of pounds contained 
in 14,400 yards of yarn. 



VOCABULARY ix 

Lea. — 300 yards of yarn. Same as cut. The number 
of leas or cuts in one pound of yarn gives the name to the 
yarn, thus yarn which measures 1800 yards to the pound 
would be called 6 lea yarn. 1800 -=-300 = 6. 

Length op Bell. — Where it is necessary to measure 
the material in the process of manufacture it is usual to 
attach a bell mechanism to some part of the machine over 
which the material passes. This mechanism is arranged 
so as to ring a bell when a certain length has been delivered. 
The length of bell is the number of yards the machine de- 
livers while the wheel which rings the bell makes one revo- 
lution. 

Loom. — The machine in which the yarn is woven into 
fabric. 

Marline. — -Coarse twine made in Mills 5 and 6. 

Polished Twine. — Twine after it is put through a 
mixture of starch, wax, and so forth, and has passed over 
rollers which put a glossy surface on it. 

Production.- — The amount of work which comes off 
one or more machines in a specified time. 

Reel. — A frame to wind yarn, called a 54", 72", or 90" 
reel according to its circumference. 

Rove. — A slightly twisted sliver. 

Roving Frame. — The machine on which the sliver is 
first formed into a thread in the process of manufacturing 
yarns. 

Sett of Laps. — A number of slivers of the same length 
whose combined weight will total a fixed number of pounds. 

Sliver. — Textile fibres formed into a continuous strand. 

Softener. — A machine having a number of fluted rollers 
pressed together with heavy springs. The fibres are passed 
between these rollers to prepare them for spinning. 

Spindle. — The vertical rotating rods on a spinning 
machine bearing the bobbins. 

Spyndle. — 14,400 yards of yarn. 

Step.— The bearing on which the spindle runs. 

Stripper. — A fast driven roller in the carding machine. 

System. — A group of machines consisting of cards, 
drawings, and roving frames. 

Tape. — Cotton webbing used in Ludlow to drive spinning 
spindles. 



x VOCABULARY 

Turns per Inch. — The number of revolutions the flyer 
makes while the rollers of a machine are delivering one 
inch of yarn. It is by the turns per inch that the twist on 
yarn or twine is calculated. 

Warp. — The threads running lengthwise in any woven 
fabric. 

Weft. — The threads running crosswise in any woven 
fabric. 

Wharve. — That part of the spindle around which passes 
the tape which causes the spindle to revolve. 

Worker. — The slow moving roller in a carding machine. 



Ludlow Textile Arithmetic 



NOTATION AND NUMERATION. 

LESSON I.— Oral. 

A Unit is a single quantity, for example, a reel, a softener, 
a pound. 

(A unit may also be a definite group of things such as 
a system, although a system is made up of several single 
machines.) 

A Number is a collection of units, for example, six reels, 
two softeners, ten pounds. 

Figures are used to represent numbers, as 

0123456789 
zero one two three four five six seven eight nine 

(The zero is sometimes called naught and cipher.) 

Notation means the method of writing numbers. 

Numeration means the method of reading numbers. 

Reading Numbers. — Numbers are read from left to 
right like ordinary printing. A single figure expresses a 
certain number of units and is said to be in the unit column. 
A number of two figures has one figure in the unit column, 
and, as the other figure represents a certain number of tens, 
it is said to be in the tens column. Each column has a 
certain name as shown below. 



IvUDLOW TEXTILE ARITHMETIC 



O 






w 
Pi 






Pi 
03 

CO 

pi 


■d 























*d 












*d 






1 

*d 


Pi 


w 


Pi 

CO 

pi 


pi 



+3 


*d 






0) 


i-H 


w 


a) 


773 


Pi 


CD 





CD 








£ 


Pi 

O 


'd 


a 


^d 


•p 


5-H 

-d 




tfl 




Pi 




Pi 


pi 


1 


Pi 

pi 


Pi 

CD 

■p 


Pi 
pi 


CD 


4-3 

'3 
pi 



397,492,608,125 

The above number would be read, three hundred ninety- 
seven billion, four hundred ninety-two million, six hundred 
eight thousand, one hundred twenty-five. 

(In reading, the name of the unit figures is omitted. 
The names of other columns, to the left, of figures of higher 
denomination, are omitted in this work.) 

Periods. — Numbers are separated into groups of three, 
by commas, beginning at the right. These groups are 
called periods. 

(As ten of any order, or column, equal one of the next 
column to the left, this method of writing numbers, is called 
the decimal system. The word decimal is derived from a 
Latin word meaning ten.) 

Writing Numbers. — To write numbers, begin at the 
left and write the hundreds, tens, and units of each period, 
putting in zeros in all vacant places, and putting in commas 
between each period and the following period. 

Like Quantities, are quantities of the same kind or 
denomination, for example, dollars, boys, bobbins, twines. 



ADDITION. 



Addition is the process of finding a number which is 
equal to two or more numbers, and the result is called their 
sum. The sign +, read plus, indicates addition. 

(Only like quantities can be added.) 



LUDLOW TEXTILE ARITHMETIC 
LESSON II.— Written. 



Add: 














1. 2. 


3. 


4. 


5. 


6. 


7. 


8. 


438 146 


400 


1400 


23,547 


376,489 


963,525 


706,145 


951 257 


584 


1276 


16,938 


976,432 


37,675 


90,676 


276 389 


224 


1342 


61,847 


43,763 


538,629 


3,865 



(Test the accuracy of the work by adding down as well 
as up.) 

9. — Find the total number of pounds in the following; 
one bale of jute weighing four hundred pounds, one bale of 
American hemp weighing five hundred eighty-four pounds, 
one bale Italian hemp weighing six hundred twenty-four 
pounds, and one bale of flax tow weighing two hundred 
twenty-four pounds. 

10. — What is the total number of pounds contained in 
the following shipment of polished twines: one bale of 
18 B American containing one hundred forty-four pounds^ 
one circular bale of seaming cord containing five hundred 
four pounds, one bale of 18 D shoe twine weighing one hun- 
dred sixty-eight pounds, one barrel of 24 BC American 
twine weighing one hundred thirty pounds, and one bag of 
36 B Italian weighing one hundred fifty-four pounds? 

11. — On March 19, 1900, the jute yarn production of No. 
34 Mill was as follows: 

12 lbs. A, Six hundred thirty -nine pounds. 

14 lbs. B, Fourteen hundred six pounds. 

14^ lbs. C, Four thousand, two hundred eighty-four 
pounds. 

15 lbs. D, Fourteen thousand, three hundred fifty-nine 
pounds. 

16 lbs. E, Eleven hundred sixty-four pounds. 

What was the total production in pounds? 

12. The following is the form used for the daily jute 
yarn production slips which are sent to the office daily at 
5 P. M. 



Department No. 38, 



LUDLOW TEXTILE ARITHMETIC 

March 18, 1902. 





JUTE 


YARN PRODUCT 


ION. 


No. 


of Bags 


Sorts 


Lbs, 




8 


10 lbs. CA 


2534 




7 


12 lbs. BB 


2478 




38 


13 lbs. BB 


14439 




35 


14 lbs. C taper 


13294 




11 


14 lbs. CC 


4269 




15 


15 lbs. BB 


5873 




5 


16 lbs. 8 Ply 


1809 


Totals, 






Over 


Find the totals. 







Note: — In the above examples the figures preceding the 
letter used in regard to the twine denotes the Size, and the 
letter denotes the Grade, while the word denotes Material 
from which the twine is made. For example, 18 B Amer- 
ican, means that the twine is called 18 size, denoting its 
diameter, is B grade, and is made from American hemp, 



JUTE YARN PRODUCTION SHEETS. 

LESSON III.— Written. 

Make out daily jute yarn production sheets or slips 
similar to the one in example 12, Lesson II, using the follow- 
ing quantities; 

1.— March. 20, 1901, the production for Dept. 67, #14 
Mill, was, 3 bags of 12 lbs. B Special weighing 798 pounds, 
5 bags of 14 lbs. B Special weighing 1726 pounds, 10 bags 
of 14^ lbs. C weighing 2604 pounds, and 62 bags of 15 C 
Special weighing 13829 pounds, and 3 bags of 16 lbs. C 
weighing 1096 pounds. Make out regular slip showing 
totals. 

2.— The production sheet of Dept. 69, Mill #14, for March 
31, 1901, was, 1 bag of 12 B Ex. S weighing 237 pounds, 4 
bags 14 B Ex. weighing 1565 pounds, 10 bags 14 C weighing 
3901 pounds, 3 bags 14 Taper weighing 1209 pounds, 10 



IvUDLOW TEXTILE ARITHMETIC 5 

bags 143^ C weighing 2557 pounds, 5 bags 15 C weighing 
2148 pounds, 31 bags 15 S weighing 6662 pounds, 2 bags 
16 C weighing 876 pounds. Find totals from slip in regular 
,orm. 

3.— The production sheet for Dept. 35, Mill #14, for April 
3, 1901, was, 6 bags 12 BX weighing 1730 lbs., 5 bags 14 BX 
weighing 1724 lbs., 10 bags of 14^ C weighing 2872 lbs., 
62 bags 15 C weighing 4988 lbs., 13 bags 16 C weighing 
5078 lbs., 2 bags 19 C weighing 660 lbs., 1 bag 24 C weighing 
484 lbs. Find totals on regular sheet. 

4. — The following sheet shows the total production of 
jute yarns in six mills for one day. Find the totals. 









April 3, 1908. 


No. of Bags 


Sorts 


Lbs. 


2 


6 lbs. 


A 


780 


8 


7 lbs. 


A 


1074 


2 


12 lbs. 


A 


540 


3 


10 lbs. 


B 


589 


8 


12 lbs. 


B 


2740 


4 


13 lbs. 


B 


1104 


17 


14 lbs. 


B 


4809 


7 


15 lbs. 


B 


1987 


14 


10 lbs. 


C 


4205 


83 


14 lbs. 


C 


23721 


70 


15 lbs. 


C 


23808 


15 


16 lbs. 


C 


5420 


2 


24 lbs. 


C 


575 


3 


30 lbs. 


C 


894 



Totals, 

Note: — Yarns tabulated like above are arranged con- 
secutively, that is, A, B, C, and so forth, 6, 7, 8, and so forth. 
In other words the yarns are arranged in the order of their 
diameters and in groups according to their grade. 

5. — Jute yarn production sheet for the mills for April 
2, 1900. Arrange in proper form, according to example 4, 
and find the totals. 5 bags of 6 lbs. AK weighing 2000 lbs., 
3 bags 7 lbs. AK weighing 954 pounds, 2 bags of 8 lbs. AK 
weighing 604 pounds, 9 bags 12 lbs. AK weighing 3475 pounds, 



6 LUDLOW TEXTILE ARITHMETIC 

3 bags." 12 lbs. KX weighing 11 84 pounds, 10 bags 13 lbs. KX 
weighing 3275 lbs., 13 bags 14 lbs..;KX weighing ;5400 lbs., 
2 bags 15 lbs. BX weighing 799 lbs.', 4 bags. 10 lbs. NX weigh- 
ing 1389 pounds, 21 bags 14 lbs. C weighing 8568 pounds, 
2 bags- 15 lbs. C weighing 8057 pounds, 10 bags 16 lbs. C 
weighing 4056 pounds, 2 bags 19 lbs. C weighing 912 pounds, 
5 b'ags 30 lbs. C weighing 2042 pounds. 

"6. — Arrange the daily total jute yarn production sheet 
and find totals for April 7, 1900. 5 bags 30 C weighing 
1741 pounds, 10 bags 16 C weighing 3575 pounds, 18 bags 
15 C weighing 5840 pounds, 14 bags 14 C weighing 6460 
pounds, 15 bags 14 MX weighing 5760 pounds, 8 bags 13 MX 
weighing 2900 pounds, 4 bags 12 AX weighing 1500 pounds, 

4 bags 10 CX weighing 1464 pounds, 4 bags 7 RX weighing 
2380 pounds, 6 bags 6 RX weighing 1370 pounds. 



PREPARING ROOM PRODUCTION SLIPS. 



LESSON IV.— Written. 

1. — Find the totals in the following Preparing Room 
Jute Yarn Production Slip. 

March 17, 1907. 
Department No. 75. 



No. of System 


Sorts 


Lbs. Laps No 


. of D. 


1* 


B Italian Twine 


3160 


18 


2 


Queen Twine 


3000 


23 


3 


Queen Twine 


2750 


22 


4 


BC American Twine 2430 


20 


5 


C Jute Yarn 


2000 


24 


6 


Springfield Baling 


2060 


18 


7 


Boston Baling 


1890 


17 


8 


C Jute Yarn 


3350 


24 


9 


BX Jute Yarn 


3500 


25 


10 


BX Jute Yarn 


2750 


26 


11 


Extra Baling 


2620 


16 



Totals, 

2. — The production sheet of Dept. # 152, for March 
20, 1905,' shows that #1 system used twenty-two hundred 
fifty pounds of laps and turned off 16 doffs of A Jute, #2 



LUDLOW TEXTILE ARITHMETIC 7 

system had two thousand one hundred fifty pounds of laps 
and 15 doffs of A "Jute, #3 system running A Jute used 
twenty-four hundred pounds of laps and turned off twenty 
doffs, #4 system had two thousand five hundred pounds 
of lap! of A Jute and turned off fourteen doffs. The daily 
slip for this day should show the sorts on each system, num- 
ber of pounds of laps used, number of doffs on each system, 
and also should have the totals for the day, made out in the 
form shown in example. 

3.-«-0n March 28, 1903, in mill #14, Dept. 60, the following 
quantities were produced: System 1 made C Jute using 
2250 lbs. of laps and turning off 24 doffs, on system #2, 20 
doffs of A Jute were turned off and 4000 lbs. of laps used, 
on system #3 running B Jute 3250 lbs. of laps were used 
and 24 doffs turned off, system #4 running C Jute used 3500 
lbs. laps and turned off 23 doffs, system #5 running C Jute 
used 3750 lbs. laps and turned off 21 doffs, svstem #6 running 
C Jute used 2500 lbs. laps and turned off 24 doffs, #7 running 
C Jute used 3050 lbs. laps and turned off 18 doffs, #8 running 
C Jute used 2450 laps and turned off 18 doffs, #9 running 
C Jute used 3750 lbs. laps and turned off 22 doffs, and system 
#10 used 2000 lbs. laps and turned off -20 doffs of C Jute. 
Make out slip and find the totals. 



WEEKLY PRODUCTION SHEETS. 

LESSON V.— Written. 

1. — If the daily total production of carpet yarn in two 
Ludlow Mills is as follows, find the total production for the 
week: 



Monday 112,475 

Tuesday 130,784 

Wednesday 107,685 

Thursday 104,746 

Friday " 109,859 

Saturday 61,846 



Check 



360,944 



297,451 



Total, 



8 LUDLOW TEXTILE ARITHMETIC 

2. — Make out slips for other weeks using the figures 
below, and find totals. 

On Monday the total production of carpet yarn was 
101,574 lbs., on Tuesday, 111,487 lbs., on Wednesday, 
115,967 lbs., Thursday, 118,243 lbs., Friday, 113,479 lbs., 
and on Saturday, 61,165 lbs. Check results as shown in 
Example 1. 

3. — The total production of carpet yarn on Monday was 
103,746 lbs., Tuesday, 110,692 lbs., Wednesday, 115,764 
lbs., Thursday, 118,765 lbs., Friday, 107,776 lbs., and on 
Saturday, 63,127 lbs. Make out a sheet in the regular 
form, finding total and checking result. 



SUBTRACTION. 



Subtraction is the process of finding the difference 
between two numbers and the result is called their differ- 
ence. The sign — , called minus, indicates subtraction. 

The Minuend is the greater number. 

The Subtrahend is the smaller number. 

(Only like quantities can be subtracted.) 

Proof. — An example in subtraction may be proved 
or tested for accuracy by adding the result, or difference, 
to the number which is subtracted or taken away, the sum 
equaling the minuend if the subtraction is performed 
correctly. 

Example. Subtract 347 from 863 and prove. 
863 minuend 
347 subtrahend 

516 difference 

863 proof 

LESSON VI.— Written. 

Find differences, and prove: 

1. 2. 3. 4. 5. 6. 7. 

849 321 8642 3084 23675 832,645 132,641 

278 219 730 2427 19498 794,658 93,867 



LUDLOW TEXTILE ARITHMETIC 9 

8. — From a lot of Italian hemp containing twelve hundred 
eighty-six pounds there are produced eight hundred twenty- 
nine pounds of polished twines. How many pounds are 
lost in manufacturing? 

9. — In 1904 the number of school children in Ludlow 
was five hundred four, in 1907 the number was six hundred 
ninety-six. What was the increase? 

10. — According to the Assessors' report for the year 1907 
there were, in Ludlow, one thousand eighty-three assessed 
polls, and from the Registrar of Voters' report for the same 
year, the number of male voters was four hundred eighteen. 
How many men paid a poll tax who were not voters? 

11. — According to the State Census of 1895 the popu- 
lation of Ludlow was 2562. At the time of the United States 
Census in 1900, the population had reached 3536. What 
was the gain in population during these five years? 

(Explain census reports.) 

12. — The total production of jute carpet yarn in the 
Mills on Wednesday, March 17, 1903, was 322,684 
lbs. Of this #18 Mill produced 100,720 lbs., and #11 Mill 
92,250 lbs. How many pounds did the other mill manu- 
facture ? 

13. — If the gross weight of a barrel of twine is 162 lbs., 
and the empty barrel weighs 19 lbs., what is the net weight? 

(Explain net and gross weight.) 

14. — It is estimated that, in 1910, when the next United 
States census is taken, the population of Springfield will 
have reached 90,000. According to the last report in 1900, 
the population was 62,059. What gain in population will 
that be in ten years? 



MULTIPLICATION. 



Multiplication is the process of taking a number a 
certain number of times or it really is a short process of 
finding the sum of a number a certain number of times, 
and the result is called the product. 

Multiplicand is the name of the number multiplied or 
the number taken a certain number of times. 



10 LUDLOW TEXTILE ARITHMETIC 

Multiplier is the name of the number which . multiplies- 
or, which shows ^how many times the multiplicand is to be 

taken. • _"„• . " "'. ' , " ; 

Abstract Numbers are those which do -not specify- 
any particular thing, for example, 6, 7, 9. 

Concrete '"Numbers indicate or refer to some particular 
thing, for example, 6 spindles, 7 boys, 9 mills. 

(The multiplier must be an abstract number while the 
multiplicand and product may both be either abstract or 
concrete. In finding the product of two abstract numbers 
either may be the multiplicand or the multiplier.) 

LESSON VII.— Written. 

1.— Multiply 267 by 2, by 3, and by 4, and then add the 
three products. 

2. — Multiply 3401 by 8 and by 9 and then add the two 
products. 

3.— Multiply 3425 by 132. 

4.— Multiply 62,084 by 4007. 

5. — If three hundred fifty pounds of Jute can be run over 
a system of preparing in one hour, how many pounds can 
be run over the same system in a week of 55 working hours ? 

6. — If the average production of each loom is 850 yards 
of bagging per day of 10 working hours, what would be the 
total production of 112 looms for the same length of time? 

7. — What would be the wages per week of 55 working 
hours of a boy who is paid at the rate of ten cents an hour? 

8. — -If one system of preparing will supply six sides of 
spinning with rove, how many sides should 16 systems 
keep going? 

(A rove is a slightly twisted sliver or fleecy strand of 
fibre.) 

9. — If the number of help required to operate a system 
of jute preparing is 7, and it requires the services of 10 girls 
in the spinning department to spin the product of this system, 
how many help would be required to operate both the pre- 
paring and spinning room in a plant running 54 systems? 

10. — If one cut or lea of yarn contains 300 yards and 48 
cuts make a spyndle, how many yards are contained in one 
spyndle ? 



LUDLOW TEXTILE ARITHMETIC 11 

11. — -Find the "number of pounds of yarn in 1450 spyndles 
of 14 lbs. C yarn. . , 

(14 lbs. yarn means that 1 spyndle weighs 14 lbs.) 

12. — Find the number of pounds of jute in a stock house 
containing 3750 bales, weighing 400 pounds each. 

(The standard weight of a bale of "jute is 400 pounds.) 

LESSON VIII.— Written! 

1.— Make out a list showing the wages paid for a week's 
work of 55 hours for each cent difference in the rate from 
5 cents per hour to 20 cents per hour. 

LESSON IX.— Written. 

1. — Make a list showing the price per ton for work that 
is paid for at the rate of from 1 to 20 cents per 100 pounds. 



DIVISION. 

Division is the process of finding how many times one 
number is contained in another. The sign of division is h-, 
read, divided by; for example 9-^3, shows that 9 is to be 
divided by 3. 

The Dividend is the number divided. 

The Divisor is the number used for dividing the dividend. 

The Quotient is the result of the division. 

The Remainder is the part left over. 

Short Division is a method of dividing when the divisor 
is small or of such a nature that the work may be performed 
mentally. 

Long Division is a method of dividing when the work 
is fully written, but is otherwise the same as short division. 

(The quotient is written over the dividend, the first 
figure being written over the right hand figure of the partial 
dividend used in obtaining it.) 

LESSON X.— Oral. 

Example. Find the quotient of 6975-^3, by short 
division. 



12 LUDLOW TEXTILE ARITHMETIC 

The divisor is written at the 
left of the dividend, and the 
3) 6975 . quotient under the dividend. 

Three is contained in 6 twice, thus 

2325 2 is the first figure in the quotient 

and is placed below the 6 of the 
dividend. Three is contained in 9 three times, in 7 two 
times with a remainder of 1. This 1 is equal to 10 of the 
next lower order and with the 5, the next figure of the divi- 
dend makes 15. Three is contained in 15 five times. This 
work would be stated, 3 in 6, 2; in 9, 3; in 7, 2; in 15, 5; 
6975 divided by 3 equals 2325. 



1.— 936 
2.— 17869 
3.-46785 
4.-786,491 
5.-369,472 



6= ? 6.-56783 

7= ? 7.— 1440 

5= ? 8.— 30456 

8= ? 9.-274568 

5= ? 10.— 896532 



100= ? 



12= ? 

11= ? 

13= ? 

15= ? 



(The work may be tested or checked by multiplying the 
quotient by the divisor and adding the remainder to the 
product. If the work is correctly performed the result 
should equal the dividend.) 



LONG DIVISION. 



Example. Divide 85323 by 23. 

3709 As 23 is more than 8 it is 

necessary to take two figures 

23) 85323 of the dividend for a partial divi- 

69 dend. Eighty-five will contain 

23, three times, but not four 

163 times, therefore 3 is the first 

161 figure in the quotient and is placed 

above the right hand figure of 

223 the trial dividend, namely 8. 
207 Subtracting 3 times 23 from 85 
leaves 16. Annexing the next 

16 remainder figure in the dividend, the next 

partial dividend becomes 163. 

This contains 23, seven times with a remainder of 2. The 

next figure of the dividend, 2, is brought down or annexed 

and it is found that this partial dividend, 22, will not con- 



LUDLOW TEXTILE ARITHMETIC 13 

tain 23 so the next figure of the dividend, 3, is brought 
down and annexed. Twenty-three goes into 223 nine times 
with a remainder of 16. Thus the quotient is 3709 and the 
remainder 16, or 23 goes into 85323, 3709 times and 16 over. 
Note: — When it is necessary to bring down more than 
one figure of the dividend, a zero should be placed in the 
quotient for each extra figure brought down. 

LESSON XI.— Written. 

Divide: 

1.-232,848 by 56. 

2.— 94,576 by 345. 

3.— 78,942 by 372. 

4.-609,725 by 25. 

5.-345,632 by 948. 

6.— 76,865 by 3425. 

7.— 89,210 by 8740. 

8.-100,456 by 3700. 

9.-372,658 by 7005. 
10.— 84,736 by 9876. 

LESSON XII.— Written. 

1. — One cut or lea of carpet yarn measures 10800 inches' 
How many yards will one lea measure? 

(12 inches make one foot, 3 feet make 1 yard.) 

2. — A spyndle of yarn contains 14400 yards. How many 
cuts does it contain? 

3. — If a thread wound once around a reel measures 90 
inches, how many inches would a thread wound around 120 
times measure? How many yards? 

4. — Now many bales of jute (400 lbs.) can be loaded into 
a car designed to carry 60000 lbs.? 

5. — The twist on yarns is calculated by the number of 
revolutions the flyer makes during the time the roller is 
delivering one inch of yarn. Under these conditions what 
would be the twist when the roller delivers two inches of 
yarn while the flyer makes 10 revolutions? 

(Express as so many turns per inch.) 

6. — Without making allowances for waste, how many 
spyndles of 30 lbs. yarn can be made from 19 bales of jute? 

(30 lbs. yarn means that 1 spyndle weighs 30 lbs.) 



14 LUDLOW TEXTILE ARITHMETIC 

7. — Find the average cuts per spindle if the total pro- 
duction is 348,900 cuts and the number of spindles in oper- 
ation is 13956. 

8. — The report of the Assessors on March 1, 1908, gives 
the total valuation of the property in the Town of Ludlow, 
subject to taxation, as $3,472,474.00. The last official 
census gives the population as 3881. From these figures 
find the per capita assessed valuation; or, in other words, 
find the average wealth of each individual in Ludlow. 



SIGNS. 

+ plus, the sign of addition. 

— minus, the sign of subtraction. 

X times, the sign of multiplication. 

-r- divided by, the sign of division. 

> greater than. 

< less than. 

.'. therefore, or hence. 



MISCELLANEOUS EXAMPLES. 

LESSON XIIL— Written. 

1. — How many pins would be required for a push bar 
first drawing frame, with two heads, each head having 30 
faller bars with 8 rows of gills on each faller, and 2 rows of 
pins in each gill with 18 pins in a row? 

2. — In a three headed 2d drawing frame there are 30 
faller bars to the head having 6 rows of gills on each bar, 
with 2 rows of pins in each gill and 20 pins in a row. How 
many pins would be required to fill the gills of one of these 
drawing frames? 

3. — Find the number of hackle pins required for a seven 
headed roving frame, with 29 faller bars in head, 8 rows of 
gills on each faller, 2 rows of pins in each gill with 13 pins 
in a row. 

4. — Find the total number of pins it would require for 
this system of preparing machinery consisting of 1st and 2d 
drawing frames and roving. 



LUDLOW TEXTILE ARITHMETIC 15 

5. — The number of watts is equal to the number of 
amperes multiplied by the number of volts. There are 
746 watts in one horse-power. How many amperes are 
required to furnish power for a 50 H. P. motor, the number 
of volts being 110? 

6. — On a twine measuring machine is an indicator which 
records the number of yards contained in one pound of 
twine. The indicator reads 182 yards at the start and 922 
yards at the finish when measuring a quantity of 9 B Amer- 
ican. What were the total number of yards? 

Find the number of yards per pound in each sort from 
the following readings of the indicator: 

Indicator 
At St; 
7.— 12 A Italian, 
8. — 18 B American, 
9. — 24 B American, 

10.— 36 B Italian, 

11.— 48 B Italian, 

12. — Triangle Baling, 

13. — Boston Baling, 

14. — Medium Cord, 

15. — A girl is paid 12^ per hundred pounds for winding 
a certain size yarn and earns $8.40 in a week; for another 
kind of yarn she is paid at the rate of 10^ per hundred pounds. 
How much more of this kind of yarn will she have to wind 
in a week to earn the same wages? 

16. — For opening jute a man is paid at the rate of 15^ 
per bale of 400 pounds. How many pounds of jute will 
he have to handle to earn $15.00? 

17. — The revolutions per minute of the main driving 
shaft in No. 12 Mill were as follows: 140, 142, 145, 141; 139, 
137, 143. What were the average number of revolutions 
per minute? 

18. — A spinning room driving shaft makes 320 R. P. M. 
How many revolutions will it make in 3 hours? 

(R. P. M. equals revolutions per minute.) 

19. — A bailer is paid at the rate of 25^ per hundred 
pounds for one kind of work; 30^ per hundred pounds for 
another kind of work; and 35^ per hundred pounds for 



At Start 


At Fini 


341 


941 


541 


921 


921 


1205 


1205 


1405 


1405 


1555 


1555 


1985 


1985 


2365 


2365 


2573 



16 LUDLOW TEXTILE ARITHMETIC 

another kind. During the week she balls 1200 pounds at 
the first rate, 1700 at the second, and 300 at the third. What 
will her wages for the week amount to ? 

20. — If one side of a dry spinning frame turns off a certain 
size yarn at the rate of 50 pounds per hour, how long will 
it take two sides to spin 6,000 pounds? 

21. — How long will it take six sides to get out the same 
order? 

22. — The following is the weight in grains of eight 50- 
yard tests of 14 C jute yarn. Find the average weight of 
the yarn? 334, 327, 346, 353, 330, 347, 350, 333. 

23.— The United States produced in 1890, 6,940,898 
bales of cotton. Of this number Texas produced 1,594,305 
bales. How many bales did the other states produce? 

24. — A lea or cut of carpet yarn measures 300 yards. 
A spyndle of yarn is 14,400 yards. How many leas in a 
spyndle ? 

25. — If 54 systems use, on an average, 432 bales of jute 
every ten working hours, find the average number of pounds 
run over each system per hour. 

26. — In 15 tons of 30 pounds C yarn, how many spyndles? 

(30 pounds indicates that yarn weighs 30 pounds per 
spyndle.) 

27. — If 7000 grains equal one pound avoirdupois, how 
many grains in one ounce? 

28. — If the net weight of jute rove on a single rove bobbin 
weighs two pounds, how many pounds of jute rove will 
there be in twenty-five doffs of a roving frame, each doff 
containing fifty-six bobbins? 

29. — A foot of lumber is 12 inches square and one inch 
thick. How many feet- of lumber are contained in a plank 
12 feet long, 1 foot wide, and 2 inches thick? 

30. — If five spinning bobbins contain two pounds of 
yarn, how many of these bobbins will have to be filled to 
make a bale of twine weighing one hundred fifty pounds? 

31.— What would be the weight of 1,000,000 pins if 42 
pins equalled one ounce? 

32. — W'hat are the wages per hour of a boy who receives 
$8.25 for a week of 55 working hours? 

33. — What wages would a man receive for opening 96 
bales of jute in one week at the rate of 15^ per bale? 



LUDLOW TEXTILE ARITHMETIC 17 

34. — Find the number of pins contained in one thousand 
pounds of card pins weighing fifty to the ounce. 

35. — Out of every 100 pounds of American hemp there 
are produced about 60 pounds of finished twines. On this 
basis how many pounds of twine could be made from a lot 
of hemp containing 10 tons? 

(1 ton equals 2000 pounds.) 

36. — How many pins would be required for the lags of 
a breaker card cylinder, each lag or stave having six rows 
of pins* with 42 pins in each row, and 150 lags being the 
usual number required to cover an ordinary card cylinder 
four feet in diameter? 

37. — What would be the weight of this number of pins 
if 42 pins weigh one ounce? 

38. — It is computed that about 12 H. P. is required to 
drive one system of preparing machinery and that about 
6 H. P. is required to drive one side of a dry spinning frame. 
On this basis what would be the H. P. required to operate 
a room containing 4 systems of preparing and 20 sides of 
spinning machinery, allowing 10 additional H. P. to drive 
shafting and belts? 

39. — In the year 1815, Catharine Woods of Dunmore,' 
near Ballynahinch, in the County of Down, Ireland, then 
about 13 years old, spun 12 cuts of linen yarn weighing 10 
grains, each cut 120 threads, each thread 2}/% yards. How 
many yards would there be in one pound of this yarn? 

(7000 grains equal one pound.) 

40. — The circumference of the earth's surface is about 
25,000 miles. A spyndle of yarn measures 14,400 yards. 
How many spyndles of yarn would it take to go around the 
earth? 

41. — How many gallons of batching oil would be required 
for a week of 55 working hours in a room where the average 
product is 30,000 pounds per 10 hours, and this oil is used 
at the rate of one gallon for each 1000 pounds of jute? 



ANALYSIS. 



Analysis is the process of reasoning from the given' 
number to one, "and then from one to the number required. 



18 LUDLOW TEXTILE ARITHMETIC 

. Example. If a weaver receives $1.00 for weaving 4 rolls of 
bagging, how much will he receive for 96 rolls? 

If $1.00 or 100^ are re- 
ceived for weaving 4 rolls of 
lOOjzf-^- 4 = 25^ bagging then as many cents as 

25^X96 = 2400^ 4 is contained times in 100 ^ 

= $24.00 or 25^ will be received for 1 
roll, and 96 times 25 cf, or 2400^ 
or $24.00 will be received for 96 rolls. 

LESSON XIV.— Written. 

1. — If a roving frame turns off 20 doffs in a day of 10 
hours, how many doffs should the same frame turn off in 
a week of 55 hours? 

2. — If a winding machine turns off 6,000 pounds of jute 
yarn in a day of 10 hours, how many pounds can be taken 
off this machine in 55 hours? 

3. — A man receives 75^ for opening 5 bales of jute, how 
much is he paid for opening 15 bales? 

4. — If 25 girls do a certain amount of work in 10 hours, 
how many girls will be required to do the same work in one 
hour ? 

5.— If 50 card pins weigh one ounce, what will be the 
weight of 10,000 pins? 

6. — If there are 2030 fallers in 10 seven-headed roving 
frames, how many fallers are contained in 16 eight-headed 
frames of the same make? 

7. — How many bales of jute will be required to make 
150 bags of yarn if 3 bales of jute make 2 bags? 

8. — With an output of 100,000,000 pounds yearly, what 
would be the average number of pounds turned out weekly? 

9. — What would the import duties amount to for 100,000,- 
000 pounds if each pound was taxed one cent? 

10. — Find the difference one eighth of a cent a pound would 
amount to in buying 100,000,000 pounds of stock. 

11. — How many feet of card clothing, 2" wide, would 
be required to cover a doffer 72" wide and 14" in diameter? 

12.— With jute selling at $112 per ton in London, what 
is it worth per pound in Ludlow after adding one cent per 
pound for cost of freight, insurance, and so forth? 

(An English ton equals 2210 pounds.) 



LUDLOW TEXTILE ARITHMETIC 19 

LESSON XV.— Written. 

1. — What is the cost per pound of hemp which sells, in 
the United States, for $245 per ton? 

(An American ton equals 2000 pounds.) 

2. — What is the cost per ton for polishing and balling 
twine if the polisher is paid at the rate of 18^ per 100 pounds 
and the bailer at the rate of 25{zf per 100 pounds? 

3. — If 4 men can open 56 bales of jute in one day of 10 
working hours, how many men will it take to open 70 bales 
in 5 hours? 

4. — Twenty-five bales of jute weigh 10,000 pounds. 
Find the weight of 17 bales. 

5. — Three winders can wind 3600 pounds of jute yarn 
in 10 hours, how many girls would be required to turn off 
12000 pounds of yarn in the same time? 

6. — How many polishing machines will have to be put 
on a certain kind of work to get out 33,000 pounds of twine 
in 55 hours if one machine can polish 1200 pounds in 10 
hours ? 

7. — How many sides of spinning will have to be put on 
to produce 99,000 pounds of yarn in .2 weeks of 55 hours 
each, if one side turns off 450 pounds of yarn in a day of 
10 hours? 

8. — How many bales of jute can be purchased for $1200, 
if one pound of jute costs 6^? 

9. — If one foot of belting costs $1.44, what would be the 
cost of a roll of belting containing 250 feet? 

10. — If 3 boys make a tool chest in 12 days, how many 
boys will be required to make it in one day? 

11. — If 7 boys do a piece of work in 8 days, how many 
boys will it take to do the same work in one day? 

12. — If 3 pounds of jute cost 27^, what will a bale con- 
taining 400 pounds cost? 



FACTORS AND MULTIPLES. 

Factors of a number are those whole numbers, which 
multiplied together, will equal the original number. 

A Prime Number has no factors except itself and one, 
for example, 2, 3, 5, 7, 11. 



20 LUDLOW TEXTILE ARITHMETIC 

Even Numbers can be divided by 2 with no remainder, 
as 2, 4, 6, 8, 10. 

Odd Numbers can not be exactly divided by 2, as 3, 5, 
7, 9, 11. 

Factoring is the process of separating a number into 
its factors, for example, the factors of 15 are 3 and 5. 

A number can be divided by 

2 if the right hand figure is even, as 246, 738. 

3 if the sum of all the figures can be divided by 3, 

as 48, 264. 
5 if the right hand figure is either zero or five, as 
35, 470. 

LESSON XVI.— Oral. 

1. — Name all the prime numbers from 1 to 51. 
2. — Name the even numbers in the following list: 

7 9 23 48 21 50 67 
5 74 26 13 36 51 53 

8 84 72 96 69 78 90 

3. — Name all the odd numbers in the above list. 
1 4. — Name the prime factors in each of the following: 
27, 36, 48, 95, 37, 108- 232, 438, 625. 

'■'■ 5. — In example 2, name all the numbers that are divisible 
by 3; by 5. 

LESSOR XVII.— Written. 

Find the prime factors of the following: 



1.— 60 


6.- 


- 75 


11.— 480 


16.- 


-1086 


2.-36 


?.- 


-210 


12.— 640 


17.- 


-1550 


3.-39 


8.- 


-400 


13.— 720 


18.- 


-3150 


4.-42 


9.- 


-810 


14.— 700 


19.- 


-3900 


5.— 17 


10.- 


-756 


15.— 800 


20.- 


-1485 



GREATEST COMMON FACTOR. 

A Common Factor of two or more numbers is a factor 
of each of the numbers. 

The Greatest Common Factor of two or more numbers 
is the greatest factor contained in each of the numbers. 

(The greatest common factor is written G. C. F. and is 
sometimes called the G. C. D. or greatest common divisor.) 



LUDLOW TEXTILE ARITHMETIC 21 

(Numbers that have no common factors are said to be 
prime to each or other, or are mutually prime.) 

Example. Find the G. C. F. of 18, 30, and 42. 

Separate the different 

numbers into their prime fac- 

Case I. 18 = 3X3X2 tors, and then take those fac- 

30 = 3X2X5 tors which are common to all 

42 = 3 X 2 X 7 three numbers and multiply 

2X3 = 6 G. C. F. them together. The product 

is the G. C. F. In this case 

the common factors are 2 

and 3, therefore 2X3 = 6 and 6 is the G. C. F. In a simple 

case of this sort the G. C. F. should be found by inspection. 

In this second method 
the numbers are arranged 
Case II. 3) 18 — 30 — 42 horizontally in a row, and 

2) 6—10 — 14 are divided by a common 

ol c _ 7 factor. The quotients are in 

9 wo = g q. p p turn divided by a common 
factor and the process con- 
tinued until the quotients 
last obtained have no factor in common. The product oj 
the divisors or common factors, in this example, 2X3 = 6, 
is the G. C. F. 

LESSON XVIIL— Written. 

Find the G. C. F. of: 

1.— 60, 72, 108. ' 6.-35, 49, 21. 

2.-54, 72, 90. 7.-75, 175, 225. 

3.-55, 99, 110. 8.-28, 77, 98. 

4.-84, 120. 9.-39, 52, 78. 

5.— 80, 112, 160. 10.— 60, 180, 240. 



LEAST COMMON MULTIPLE. 

A Multiple of a number is a number which will exactly 
contain the first. 

A Common Multiple of two or more numbers is a number 
which will contain each of the numbers without remainder. 



22 UJDLOW TEXTILE ARITHMETIC 

The Least Common Multiple of two or more numbers 
is the smallest number that exactly contains all of the given 
numbers. 

(Least common multiple is written L. C. M.) 

LESSON XIX.— Oral. 

Name the least common multiple of: 

1.— 2 and 3. 

2. — 5 and 6. 

3.-7 and 8. 

4.-9 and 4. 

5.-2, 3, and 4. 

6.-2, 4, and 5. 

7.-3, 6, and 10. 

8.-8, 6, and 12. 

9.-2, 4, and 10. 

10.— 9, 3, and 4. 

LESSON XX.— Written. 

Example. Find the L. C. M. of 54, 72, 90. 

Any multiple of 54, 
72, 90, will contain all 
Case I. 54 = 2X3X3X3 . the factors of each of the 

72=2X2X2X3X3 three numbers. There- 

90=2X3X3X5 fore, to find the least 

2X3X3X3X2X2X5 common multiple, it is 
= 1080 L. C. M. necessary to find the 

product of all the prime 
numbers, or factors, us- 
ing each factor the greatest number of times that it occurs in 
any one number. 

This second method 
is usually more conven- 
Case II. 2)54-72-90 ient. Arrange the num- 

bers in a line and divide 
by any factors that are 
, common to any two of 

3-4-5 them. When the last 

2X3X3X3X4XO quotients are prime to 

= 1080 L. C. M. each other, multiply 

them together and then 
multiply this product by each of the divisors. The final 
product of them all is the L. C, M. 



2) 


54- 


-72- 


-90 


3) 


27- 


-36- 


-45 


3) 


9- 


-12- 


-15 



LUDLOW TEXTILE ARITHMETIC 23 



LESSON XXI.— Written. 



Find the L. C. M. of 
1.— 9, 36, 45. 
2.-6, 33, 99r 
3.-35, 70, 140. 
4.— 15, 30, 45, 90. 
5.— 12, 48, 36. 
6.— 16, 48, 96. 
7.-4, 12, 18. 
8.-*9, 15, 45. 
9.— 300, 400, 500. 
10.— 20, 16, 8. 



CANCELLATION. 



Cancellation is a short process in division, whereby- 
equal factors may be removed from both divisor and dividend. 
This process is used in calculating drafts and twists in mill 
work, and in tracing speed, or R. P. M. from one part of a 
machine to another. 

10^-5 means that 10 is to be divided by 5. This also 
may be expressed 10/5. 

By short division this would be worked as follows: 
5) 10 
2 
Thus 10 over 5 or 10 divided by 5 equals 2. 
A shorter way of finding the quotient would be as follows : 
2 5 is an equal factor of both 10 

10 2 and 5 and goes into 10 twice and 5 

— =— = 2 once. As 1 goes into 2 twice the 

5 1 final result is 2. This process is 

1 called cancellation. 

Example. Divide 9 X 18 X 24 by 3 X 6 X 12. 
1 
3 6,4 
9X18X24 = 3X6X1 

= =.18 

3X 6X12 1X1X1 
1 1 8 
1 
(By this method it is not necessary to multiply the differ- 
ent quantities together before dividing.) 



24 LUDLOW TEXTILE ARITHMETIC 

LESSON XXII.— Written. 

Divide : 

1.—7X6X16 by 3X8x14. 
2.— 11X27X30 by 9X15X3. 
3.-9 X 14 X39 by 13 X7 X 18. 
4.— 1728X3X7 by 12X12X12. 
5.— 84X13X5 by 91X4X15. 



UNITED STATES MONEY. 

In the Ludlow Mills the cost of material, and cost of 
labor and manufacture is reckoned in dollars, cents and mills. 
The relationship of these different units is shown in the 
following table: 

10 mills — 1 cent (^) 
100 cents = 1 dollar ($) 

(The dollar sign, $, is written before, or at the left of the 
figures. Thus twenty-four dollars would be written $24. 
Twenty-five cents would be written 25$/, the sign, 0, following 
the figures. When both dollars and cents are written to- 
gether, a period or point is used to separate one from the 
other. This period is called the decimal point. The dollar 
sign only, is used. Thus $24.25 means twenty-four dollars 
and twenty-five cents. Mills are written in the third decimal 
place. Thus $9,467 would be read nine dollars and forty- 
six cents, seven mills.) 

Day workers are those who are paid a stated sum per 
week. Their rate per hour is found by dividing 55 into their 
wages for a week. 

Piece workers are those who are paid a certain rate for 
performing a certain amount of work, as, for instance, a 
reeler may receive 6 cents for every reel of yarn she turns off 
her machine, or a man may receive 15 cents for every bale 
of jute he opens. 

In making up the wages of both day and piece workers, 
parts of a cent are used. These parts are called mills. 

LESSON XXIIL— Written. 

Write the following in figures: 

1. — Ten thousand dollars and forty cents. 



LUDLOW TEXTILE ARITHMETIC 25 

2. — Seven hundred eighty-four dollars and sixteen cents, 
four mills. 

3. — Five thousand six hundred seventy-four dollars and 
nineteen cents, two mills. 

4. — Three million dollars and three mills. 

5. — Seven thousand dollars and three cents, three mills. 

6. — Eight hundred seven dollars and ten cents. 

7. — Five thousand twenty-three dollars and seventy-six 
cents, eight mills. 

8.-»One dollar and two mills. 

9. — Twenty-seven dollars and thirty cents, five mills. 

(Use ciphers to fill in vacant places.) 

TO REDUCE DOLLARS AND CENTS TO CENTS. 

Example. Reduce $1.27 to cents. 

Dollars and cents are reduced 
to cents by dropping the decimal 
$1.27 — 127^ point and substituting the sign for 

cents, i, for the dollar sign. 

(Dollars, cents, and mills may be reduced to mills in the 
same way.) 

LESSON XXIV.— Written. 

1.— Reduce $22.46 to cents. 

2. — Reduce $47 to cents. 

(If no cents are expressed annex two zeros.) 

3 — Reduce $6,759 to mills. 

4.— Reduce $32,476 to mills. 

5 —Reduce $66,432 to mills. 

TO REDUCE CENTS TO DOLLARS. 
Example. Reduce 3768^ to dollars. 

Cents are reduced to dollars 

and cents by pointing off two 

3768<z( — $37.68 decimal places from the right 

and substituting the dollar sign 

for the sign for cents. 

(If mills are to be reduced to dollars, point off three places 

and proceed as when changing cents to dollars.) 

LESSON XXV.— Written 

Reduce : 

1.-4987^ to dollars. 

2. — 7658 mills to dollars. 



26 LUDIvOW TEXTILE ARITHMETIC 

3—46036^ to dollars. 

4.— 9000145 mills to dollars. 

5.— 10049015 mills to dollars. 

LESSON XXVI.— Written. 

What would be the wages per week of 55 hours for a boy 
who is paid at the rate of 

1. — 8^ 6 mills per hour? 

2. — 9^ 1 mill per hour? 

3. — 9^ 5 mills per hour? 

4. — 10^ 5 mills per hour? 

5. — 10 s^ 9 mills per hour? 

6. — 11^ 4 mills per hour? 

7. — 120 7 mills per hour? 

The word Mills is derived from the Latin word meaning 
one thousand, and is used because 1000 mills equal 1 dollar, 
the latter being the unit of computation. 

The word Cent is from the Latin word meaning one 
hundred, and is used because 100 cents equal 1 dollar. 

Dollar is a word taken from the German word, thaler, 
a money unit first coined in Germany. 



DECIMAL FRACTIONS. 



A Fraction is a part of a whole thing or unit. There 
are two general classes of fractions known as common or 
vulgar fractions and decimal fractions. As decimal fractions, 
or decimals, are used for practically all calculations in mill 
work, a very definite and complete knowledge of them ought 
to be obtained. Common fractions are considered more 
fully in the next chapter. 

From information obtained in the chapter on United 
States money we know that: 

1^ may be written $0.01, and is really one hundredth 
part of a dollar. 

The period, or point, is called the decimal point. 

10^ may be written $0.10, and is one tenth part of a 
dollar. It is likewise ten hundredths of a dollar. 



LUDLOW TEXTILE ARITHMETIC 27 

25^, 5 mills, may be written $0,255. If 25^, 5 mills are 
reduced to mills, the result will be 255 mills or two hundred 
fifty-five thousandths of a dollar as one mill is one thousandth 
part of a dollar. 

One dollar and fifteen cents may be written $1 . 15. 

Thus it may be seen that, when whole dollars are con- 
sidered, they are always placed at the Left of the decimal 
point and that, when parts of a dollar are considered, they 
are alw*ays placed at the Right of the point. All figures in 
the first column to the right are tenth parts of a dollar while 
all figures in the second column to the right are hundredths 
and all figures in the third column are thousandths of a 
dollar. 

Figures at the left have the same relative value and each 
column has the same name as ordinary numbers, and ten 
of one column will equal one of the next column to the left. 

Likewise with figures at the right of the decimal point, 
ten of one column will equal one of the next column to the 
left of it. 

This system of representing whole numbers and frac- 
tional parts is called the Decimal System. It is evident 
that it may be applied to any other unit as well as to dollars. 

In reading decimals the word And is usually substituted 
for the decimal point. The decimal is read as if a whole 
number with the name of the column of the last figure at the 
right. 

Example 1. Read 4.37. 

4.37 would be read four and thirty-seven hundredths. 

Example 2. Read 325.0046. 

325.0046 would be read three hundred twenty-five and 
forty-six ten thousandths. 

Example 3. Read 0. 1038. 

. 1038 would be read one thousand thirty-eight ten 
thousandths. 

( When there is no whole number the zero is usually 
placed at the left of the point and the word, and, is omitted 
in reading. ) 



28 LUDLOW TEXTILE ARITHMETIC 

NUMERATION TABLE. 



-d -a 

£ w £ 

*d 5 'd _ ij 2 5 -d r3 



*d 




p! 




a5 


w 


o 


13 


+J 


Pi 


pj 


O 


<u 


,£j 


H 


H 



■2 ^ ^ * ^ w £ .§ £ ^ ? ^ ^ .2 

g W Eh ^£h DQHffi Eh H K ^ 
3 517 28 6.41 96 37 
7th 6th 5th 4th 3rd 2nd 1st 1st 2nd 3rd 4th 5th 6th 



05 


PI 

03 


*d 
pi 

o3 
w 


Pi 
O 
,3 


pi 


+s 


O 


p| 


-O 


cu 


H 


H 



LESSON XXVII.— Oral. 



Read: 
1. 



1.- 


-37.46 


6.- 


— 4.372 


11.- 


- 7.4396 


2.- 


—42 . 02 


7.- 


— 6.295 


12.- 


- 3.4005 


3.- 


-29.08 


8.- 


- 8.001 


13.- 


- 7.04396 


4.- 


—25.10 


9.- 


- 0.103 


14.- 


—10.00025 


5.- 


- 0.39 


10.- 


-10.010 


15.- 


—46 . 29003 



LESSON XXVIII.— Written. 

Write in figures: 

1 — One tenth. 

2. — Twenty-seven hundredths. 

3. — Four hundred thirty-seven thousandths. 

4. — Six hundred eight and two thousand seven hundred 
one ten thousandths. 

5. — Nine hundred forty-one and thirty-two thousand 
seven hundred twenty-one hundred thousandths. 

6. — Three and forty-one thousand one hundred fifty- 
nine hundred thousandths. 

7. — Nineteen and eighty-three thousandths. 

8. — One thousandth. 

9. — One thousand seven hundred sixty-three hundred 
thousandths. 

10. — Three thousand three hundred thousandths. 
11. — Sixty and six thousandths. 

12. — One hundred seventy and seven hundred four thous- 
andths. 



LUDLOW TEXTILE ARITHMETIC 29 

13. — Five and eight hundred fifty-six ten thousandths. 
14. — Eighty-nine and nine, thousandths. 
15. — Three hundred seventy-eight ten thousandths. 
16. — Five and nine thousand seven hundred sixty-four 
ten thousandths. 



ADDITION OF DECIMAL FRACTIONS. 

Addition of decimals is performed in exactly the same 
manner as with whole numbers, care being taken to place 
like columns under each other. 

Example. Add 13.27, 46.09, 55.38, 94.573, 10.40765, 
32 . 00469. 

If the decimal points are arranged 
13.27 in a vertical column, then like columns 

46.09 will also come in line. Each column 

55.38 is added as with whole numbers, in 

94.573 every case ten of one column being 

10.40765 equal to one of the next column. The 

32 . 00469 decimal point in the sum should be 

■ placed directly below the other deci- 

251.72534 mal points. 

LESSON XXIX.— Written. 

Find the sum of: 

1.— 36.16, 45.76, 87.11. 

2.— 143.45, 76.84; 60.176. 

3.— 145.342, 47.0086, 32.608. 

4.— 75.8069, 90.7165, 22.76405. 

5.— 48.78506, 401.57246, 62.004709. 

6.— 47.7604, 0.41647, 11.479205. 

7.— 71.47605, 18.4875309, 0.143057. 

8.— 49.51605, 43.1431, 56.48523. 

9—84.875604, 32.004305, 57.49987. 
10.— 12.0047, 57.647008, 0.7415645. 
ir— 99.8577, 44.53423, 40.1457049. 
12.— 40.1411, 57.776405, 21.432. 

13. — Find the sum of two dollars and seventy-three 
cents, one hundred fifty-nine dollars and six cents, twenty-two 
dollars and fifty-four cents, one thousand four dollars and 
nineteen cents. 



30 LUDLOW TEXTILE ARITHMETIC 

14. — Add together sixty-four cents, seven dollars and a 
quarter, one hundred dollars and forty cents, and five dollars 
and ten cents. 



SUBTRACTION OF DECIMAL FRACTIONS. 

Example. Subtract 8 . 7604 from 18 . 3927. 

Place the decimals in such a way 

18.3927 that the decimal points will be directly 

8.7604 under each other. Subtract as if the 

figures were whole numbers and place 

9.6323 the decimal point in the remainder 

directly under the other decimal points. 
( As one tenth is equal to ten hundredths, shown by the 
ten cent piece which is one tenth of a dollar and also ten 
hundredths of a dollar, the decimal representing the same 
may be written . 1 or .10. Annexing a zero to any decimal 
does not change its value.) 

Example. Subtract 5 . 3426 from 74 . 36. 

74.3600 As a zero annexed to a decimal 

5 . 3426 fraction does not change its value 

annex two zeros to the minuend and 



69.0174 proceed in the usual manner. 

LESSON XXX.— Written. 

Find the difference between: 

1.— 3. 746 and 0.367. 

2.-6.36 and 0.1364. 

3.-67.783 and 62.999. 

4.— 0.0067 and 0.00076. 

5.— 5. 7368 and 0.94605. 

6.— 32.0001 and 26.05. 

7.-84.5768 and 54.70047. 

8.-25 and . 0025. 

9.— 100 and 0.01. 
10.— 38 and 4.607. 
11.— 10.765 and 9.1167. 
12.— 78.86 and 62.00165. 
13.— 0.0035 and 0.000789. 
14.— 23. 0016 and 17.009. 



LUDLOW TEXTILE ARITHMETIC 31 

MULTIPLICATION OF DECIMAL FRACTIONS. 

Example. Multiply $0.25 by 10. 

10X25^ = 12.50, or 10X$0.25 = 

$2 . 50. The same product is obtained 

$0 . 25 by moving the decimal point one place 

10 to the right. Any decimal may be mul- 

tiplied by 10 by simply moving the point 

$2 . 50 one place to the right. As 100 = 10 X 10, 

to multiply a decimal by 100, move the 
point two places to the right, and so on. 

Rule. To multiply a decimal by 10 move the decimal 
point one place to the right; by 100, two places; by 1000, 
three places, and so on. Annex zeros if necessary. 

Example. Multiply 50 by 0. 1. 

Multiplying by 0.1 is really finding 
50 one tenth part and one tenth of 50 is 5. 

0.1 As 10^ is one tenth part of a dollar and 

one tenth of lOjzf is one cent or one hun- 

5.0 dredth part of a dollar, so, in all cases, 

one tenth of one tenth is one hundredth. 
Thus to multiply any number by 0.01 is the same as multi- 
plying by 0. 1 twice, and to do that simply move the decimal 
place, or point, two places to the left. 

Rule. To multiply by 0.1, move the decimal point one 
place to the left; by 0.01, two places; by 0.001, three 
places, and so on. 

Example. Multiply 3 . 64 by 2. 

2 times 4 hundredths equals 8 hun- 
dredths; 2 times 6 tenths equals 12 
3.64 tenths which equal one whole one and 

2 two tenths over; 2 times 3 equals 6, 

which, added to the whole one equals 7; 
7.28 or 2 times 3.64 equals seven whole ones, 

2 tenths, and 8 hundredths, or 2 X 3 . 64 = 
7 . 28. The product has as many decimal 
places as are contained in both multiplier and multiplicand 
in this case, two. 



32 LUDLOW TEXTILE ARITHMETIC 

Example. Multiply 2 . 35 by . 4. 

To multiply by 0.1, it would be 

necessary to merely change the decimal 

2.35 point one place to the left, making the 

0.4 product 0.235. As 0.4 is 4X0.1, it 

would be necessary to multiply this 

0.940 product, 0.235 by 4, giving 0.940 for 

the final product. The final product 
has as many decimal places as are con- 
tained in both multiplier and multiplicand. 

Rule. To multiply decimal fractions, proceed in the 
regular way as with whole numbers, and then point off, in 
the product, as many decimal places as are contained in both 
multiplier and multiplicand. 



LESSON XXXI.— Written. 

Multiply: 

1.— 2 . 46 by 17. 6.-39 . 4 by 4 . 642. 

2.— 25 by 3.4. 7.— 3. 1416 by 9. 

3.— 16. 2 by 7.5. 8.— 5000 by 2. 1983. 

4.— 16 . 2 by . 75. 9.-6 . 29 by 1 . 001. 

5.— 16 . 2 by . 075. 10.— 78 . 45 by . 0026. 

11. — If the average yield of finished twine is 75 lbs. from 
each 100 lb. of hemp, how many pound balls of twine would 
be produced from one ton of hemp containing 2000 lb. ? 

12. — -The circumference of a roller is 3.1416 times its 
diameter. Find the circumference of a spinning roller whose 
diameter is 4.25". 

13. — -If jute is worth $115.75 per ton, what would be the 
value of a cargo of jute containing 6000 tons? 

14. — If a spinner is paid at the rate of 16.4 cents per 
hour, what would be her wages for a week of 55 hours? 

15. — If a boy receives 8.6 cents per hour, what are his 
wages for 25 hours work? 

16. — The diameter of the drum on a lapping machine is 
22.25 inches. How many inches of sliver passes under the 
drum of the machine in 12 revolutions? 



LUDLOW TEXTILE ARITHMETIC 33 

DIVISION OF DECIMAL FRACTIONS. 

Example. Divide '$3 . 57 by 7. 

7 will not go into 3 whole dollars, 

therefore these 3 units may be changed 

7 ) $3 . 57 to 30 tenths. Adding the 5 tenths 

makes 35 tenths. 7 will go into 35 

$0.51 tenths 5 tenths times, that is, dividing 

35 ten cent pieces by 7 will give 5 ten 

cent pieces. 7 will go in 7 one cent 

pieces once, or 7 will be contained in $3.57 just $0.51. The 

decimal point in the quotient is directly beneath ( or above ) 

the decimal point in the dividend. 

Example. Divide 3.57 by 0.7 

Multiply both divisor and dividend 

0.7) 3 . 57 by ten ( move the decimal points one 

place to the right ) . This does not 

Move decimal change the relative values of divisor 

points one place to and dividend. As the divisor is now 

right. a whole number proceed as above as 

07 . ) 35 . 7 it is evident that the concrete value 

■ of the unit does not affect the method. 

5.1 

Rule. Make the divisor a whole number by moving the 
decimal point to the right and move the decimal point of the 
dividend as many places to the right as it was moved in the 
divisor. Proceed as with whole numbers, placing the decimal 
point in the quotient in the same vertical column as it is in 
the dividend when the division is performed. 

( As multiplying both dividend and divisor by the same 
number, moving the decimal points to the right, does not 
change the relative values, so dividing both divisor and divi- 
dend by the same number, moving the decimal points to the 
left, does not change the relative values. Thus, when the 
divisor ends in one or more zeros, cross off the zeros and move 
the decimal point of the dividend as many places to the left 
as zeros were eliminated). 

( When no decimal point is expressed it is understood to 
be at the right of the unit column ) . 



34 LUDLOW TEXTILE ARITHMETIC 

LESSON XXXII.— Written. 

Find the quotient of: 

1.— 35.674-4.32. 11.-4.89-^2.44 

..2.— 72. 85 -=-5. 45. 12.-36^-6000. 

I 3.— 29. 89 -=-6. 72. - 13.— 87.265-0.292. 

4.— 2.38-0.49. 14.-130.4-5000. 

5.— 67.2-3.04. 15.— 287.9 -r- 4. 69. 

6.-17.89-^47.3. 16.— 75-300. 

7.— 7.33-5.08. 17.— 100-12.5. 

< 8.— 4.81-500. 18.— 100-16.66. 

9.— 6.04-2.32. 19.-1523.4-15,000. 

( 10.^-0.375^-250. 20.-2129.8-^21,000. 

21. — A boy is paid $4.75 for a week's work of 55 hours. 
What is his rate per hour? 

22. — An English sovereign is worth in United States 
money, -$4.86. There are 20 shillings in a sovereign. Find 
the value of an English shilling in United States money ? 
: 23. — Find the diameter of a reel whose circumference is 
90",? 
■I 24.— Find the diameter of a 54" reel? Of a 72" reel? 

25. — The circumference of the fluted rollers used in wet 
spinning is about 3.4 times their diameter. Find the diame- 
ter of a fluted roller whose circumference is 12.75"? 

26. — -Find the diameter of a reel to measure one yard of 
yarn each revolution ? 

'27. — If the speed of a roller is 25.4 R. P. M., how many 
minutes will it take this roller to make 1000 revolutions? 

28. — -The circumference of an ordinary reel is 2.5 yards. 
llow many turns will it have to make to reel one cut equal to 
300 yards"? 



REDUCTION OF COMMON FRACTIONS TO DECIMAL 
FRACTIONS. 

A Common Fraction is an expression containing two num- 
bers, one written above the other with a horizontal line be- 
tween them. 

( For a better and more complete definition of common 
fraction see chapter on Fractions .) 



LUDLOW TEXTILE ARITHMETIC 35 

As stated in chapter on Cancellation, an expression in 
this form, one number over another with a line between, 
may be considered as an expression of division where the 
upper number is the dividend and the lower number is the 
divisor. A 

Example. Reduce 3/5 to a decimal fraction. 

5.) 3.0 5 will not go into 3. Therefore 

change the 3 to tenths, then 5 will go 

0.6 into 30 tenths, 6 tenths times. 

Therefore 3/5 = 0.6. ;'.j 

Example. Reduce 23/64 to a decimal. 

64 ) 23 . 00000 ( 00 . 359375 Proceed as in the divi- 

192 sion of decimals.,., If 



there is a remainder 

380 greater than half, , the 

320 divisor then the last 

figure in the quotient 

600 should be increased by 

576 1. In either case the 

— ■ quotient would not be 

240 exact but it would be 
192 nearer by making the in- 
crease. , \ 



480 
448 



320 
320 




Rule. Divide the upper figure by the lower. 

LESSON XXXIII.— Oral. 

Change to decimals. 

1. — 1/2. 2.— 1/4. 3.— 1/5. 4.— 1/10. 5.-2/5. 

6.— 3/10. 7.-3/4. 8.— 5/10. 9.— 1/8. 10.— 1/6. • 

LESSON XXXIV.— Written. 

Change to decimals. 
1.— 3/8. 
2.— 7/16. 
3.— 9/16. 



36 IvUdlow textile; arithmetic 

4.-5/8. 

5.— 11/16. 

6.-5/32. 

7.-7/8. 

8.-9/32. 

9.— 11/64. 
10.— 25/64. 
II— 1 3/16. 
12.— 5 3/4. 
13.— 11 13/32. 
14.— 40 29/64. 
15.— 18 15/32. 
16.— 9 63/64. 
17.— 8 11/25. 
18.— 2 19/48. 
19.— 39/47. 
20.— 77/79. 



DECIMALS. MISCELLANEOUS EXAMPLES. 

LESSON XXXV.— Written. 

1. — In the process of manufacturing some grades of 
American hemp into yarn, there is a loss by waste of 0.35 
lb. out of every pound of stock. Find the loss, by waste, 
from 2000 lbs. of this particular kind of hemp ? 

2. — A micrometer is marked in one thousandths parts of 
an inch. At what place, decimal, should it be set to measure 
a 5/8" screw? 

3. — A card cylinder is 48" in diameter and is running at the 
rate of 210 R. P. M, Find the surface speed of this cylin- 
der in feet per minute ? 

(Circumference equals diameter times 3.1416. ) 

4. — Express decimally, the difference in the circumferences 
of two rollers, one measuring 4" in diameter, the other 3 15/16". 

5.— A girl receives 11.5 cents for winding 75 pounds of 
twine, and for jute yarn is paid at the rate of 7.7 cents per 
100 pounds. In a days work she winds 600 pounds of each. 
What are her wages for the day? • 

6. — -A roving frame roller is 2 . 25 inches in diameter. 
How many revolutions will it make in delivering 100 yards 
of rnve ? 



LUDLOW TEXTILE ARITHMETIC 37 

7. — How many lags would be required to cover the doffing 
roller of a card 6 feet wide ( across the face ) and 15 . 3 inches 
in diameter, if the lags are 24" long and 3" wide? 

( Lags are laid on cylinder with length in direction of 
width of cylinder. ) 

8. — -A boy is paid at the rate of 8.6 cents an hour for his 
work in the mills and he receives 1/2 that rate for the hours 
that he attends the Textile School. What would his wages 
amount to if he worked 20 hours and attended school 15 
hours?" 

9. — If the price paid for opening jute is 14.5 cents per 
bale and 4.2 cents per bale for softening, what are the wages 
a man earns who opens and softens 12 bales of jute in one 
day? 

10. — A girl is paid for her work on A yarn at the rate of 
7.7 cents for 100 pounds, B yarn at the rate of 6 . 5 cents per 
100 pounds, and C yarn at the rate of 7 . 3 cents per 100 
pounds. How much money will she have earned after mak- 
ing 456 pounds of A, 545 pounds of B, and 894 lbs. of C? 

11. — In making a sample of twine from 100 pounds of 
flax, there is a loss, by waste, in the spinning and preparing 
of 15.25 lbs., in the twisting and polishing 6.50 lbs. How 
much twine was made out of the sample ? 

12. — Running on a fifty-four hours schedule, find the 
rate per hour for work that is paid for at the rate of $6.00 
per week. 

13. — If yarn weighing one pound per spyndle equals 
24.3 grains, per 50 yards, what would 50 yards of 10 lbs. 
yarn weigh? 

14. — What would be the weight of 100 spyndles of yarn 
if each spyndle weighed 6 . 75 lbs ? 

15. — Find the cost of a gross of bobbins at 3.25 cents 
each. 

16.— What would it cost to weigh 100,000,000 lbs of 
stock if the average cost for weighing one ton of 2000 lbs. is 
14.5 cents? 

17. — If a carload of yarn weighing 60,000 pounds is worth 
$6,900 what is the yarn worth per pound? 

18. — How many reels averaging 37.5 lbs. to the reel 
would be contained in a shipment of twine amounting to 
15,000 pounds 



38 LUDLOW TEXTILE ARITHMETIC 

19. — -If the net weight of yarn on a spinning bobbin (full) 
is equal to 0. 6 of a pound, how many of these bobbins would 
be required to fill a box containing 190 lbs. of yarn? 

20.- — How many cuts in 17.5 spyndles? 

21. — How many cuts in 804.6 spyndles of A yarn? 

22. — How many cuts in 302.3 spyndles of B yarn? 

23. — How many cuts in 208.6 spyndles of C yarn? 

24. — If the loss in manufacture of twine is . 33 from 
American hemp and 0.18 from jute, how much more twine 
would be made from 100 lbs. of hemp than from 75 lbs. of jute? 

25. — -If it requires 5.5 H. P. to drive one side of spinning 
machinery, how many H. P. are necessary to drive a room 
containing 61 sides of spinning? 

26. — What would be the total H. P. required to drive, 
8 cards, 8 drawings, and 4 rovings, if a card requires 2.75 
H. P., roving 3.00 H. P. and drawings 0.625 H. P.? 

27. — If 100 yards of rove weigh 8.7 ounces, what would 
10 yards weigh? 

28. — If a spinning frame containing 156 spindles is mak- 
ing yarn at the rate of 429 cuts per hour what are the average 
number of cuts per spindle? 

29. — The cylinder of a card is- 48" in diameter, how many 
R. P. M. will it have to be driven to attain a surface speed 
of 2513.28 feet per minute? 

30. — The wheel of a bicycle is 28" in diameter, how many 
revolutions will it make if ridden from here to the Springfield 
Post Office, a distance of 7.8 miles? 



PROPORTION. 



Simple proportion, on account of its great utility and 
extensive application to mill work, is of the utmost import- 
ance in this branch of the textile industry, and it is essential 
that the general rules of proportion be thoroughly mastered 
as early as possible. In determining the gears to be used 
on the different machines and in changing from one kind of 
work to another the rules of proportion are used extensively. 

A Ratio is the relation between two numbers of the 
same kind, or, stating it in another way, in comparing two 
quantities of the same kind, the quotient obtained by divid- 
ing the first by the second is called the Ratio of the first to 
the second. 



LUDLOW TEXTILE ARITHMETIC 39 

Thus the ratio of 27 to 9 is 3, or that of 3 to 1. 

( As like quantities, only, may be compared or divided, 
the ratio is always an abstract one, so abstract numbers may 
be used in place of concrete ones.) 

i LESSON XXXVI— Oral. 

1. — Separate 15 in the ratio of 2 to 1. 

2. — What is the ratio of the number of days spent in 
school to the number of days of the week? 

3. — What is the daily ratio of the number of hours of 
school "work to the number of hours of mill work? 

4. — If there are 12 boys in the A Class and 9 in the B 
Class, what is the ratio of boys in the latter to the number, 
in the former? 

5. — Two numbers are in the ratio of 4 to 1. The smaller 
is 25, what is the larger? 

Proportion is the equality of two ratios, for example, 
10 is double 5 and 12 is double 6, therefore the ratio of 10 
to 5 equals the ratio of 12 to 6. 

This would be expressed: 

10 : 5 : : 12 : 6 
and is called, in this form, a proportion. 

Each part or number is called a Term. Thus a propor- 
tion has four terms. The proportion, as written would be 
read 10 is to 5 as 12 is 6. 

When the proportion is correctly written the product of 
the two outside terms is equal to the product of the two 
inside terms. 

Thus 6 X 10 = 5 X 12 for both = 60. 

This fact may be taken advantage of in finding the fourth 
term when only three are known. 

Example. 10 : 5 : : 12 : ? 
5 X 12 = 10 X ? 
5X12 = 60 
60 -f- 10 = 6 = the missing term. 

The above operation could be combined and the result 
obtained by cancellation. 

Thus: 1 6 

hxii 

= 6 



40 LUDLOW TEXTILE ARITHMETIC 

Rule. Arrange the three terms or numbers in the same 
line, in succession, placing the one which is of the same kind 
as the required term the Third in order, and, if, by. nature 
of the question, the required term is to be greater than the 
third term, put the greater of the two other terms in the 
Second place, otherwise put the less in that place. Then 
find the product of the second and third terms and divide 
by the First. The quotient will be the fourth proportional. 

LESSON XXXVII.— Writen. 

Example. What draft pinion would be required to 
make yarn weighing 230 grains to the 50 yards if a 38 tooth 
pinion is required to make yarn weighing 175 grains to the 
50 yards? 

As the fourth term is 
to be the number of teeth 
175 : 230 : : 38 : ? required, 38, the number 
230X38 = 8740 of teeth known must be 

8740-^175=49.99 the third term. As yarn 

Answer 50 teeth of 230 grains will require 

a larger pinion then the 

larger of the two other 

terms must -occupy the 

second place. 

(In this example 50 is the nearest whole number to the 

result obtained, consequently the gear desired would have 

50 teeth.) 

1. — If a rove weighing 9 ounces to the 100 yards is made 
into yarn weighing 14 pounds per spyndle, what would be 
the weight per spyndle of yarn made from rove weighing 10 
ounces per 100 yards, with the same draft pinion on the 
spinning frame ? 

2. — If a rove weighing 5^ ounces per 100 yards pro- 
duces yarn weighing 120 grains to the 50 yards, what would 
be the weight of yarn produced with the same gears from a 
6 ounce rove? 

3. — If a 40 draft wheel is producing a draft of 10 on a 
roving frame, what wheel would be required to produce a 
draft of 8? 

(Examples like this may be solved by analysis.) 
4. — If a 36 draft wheel on a spinning frame is m'aking 
14 lbs. of yarn what wheel will be .required to spin 12 lbs. 
yarn from the same weight of rove? 



LUDLOW TEXTILE ARITHMETIC 41 

(On a dry spinning frame the smaller the draft wheel the 
longer the draft.) 

5. — If a 50 draft wheel on a spinning frame is making 
14 lbs. yarn, what wheel will be required to spin 12 lbs. yarn 
from the same weight of rove ? 

6. — If a 14 inch pulley on the cylinder of a 4^ inch 
dry spinning frame is driving the spindles at the rate of 
1760 R. P. M., at what rate of speed will the spindles travel 
if the pulleys are changed to 16 inches? 

7: — If 10 men open 156 bales of jute in 10 hours, how 
many bales of jute will the same number of men open in 9 
hours ? 

LESSON XXXVIII.— Written. 

1. — A polisher can finish 964 lbs. of a certain size twine 
in a day of 10 working hours, how much twine should be 
taken off this machine in a week of 55 hours with 3 hours off 
for cleaning the machine? 

2. — An automatic tow feeder which has 20 inch pulleys 
is required to be driven at the rate of 80 R. P. M., what size 
drum on the driving shaft will be required to produce this 
speed, the driving shaft being driven at the rate of 267 R. 
P.M.? 

3. — A spyndle of yarn (14,400 yards) weighs one pound 
and one pound equals 7000 grains. Find the weight, in 
grains, of 50 yards of the same yarn. 

4. — The size of dry spun yarn is determined by the weight 
per spyndle. If yarn weighing one pound to the spyndle 
equals 24 . 3 grains to the 50 yards, what would be the weight 
of 50 yards of yarn weighing 14 pounds to the spyndle? 

Make out a table showing the weight in grains of 50 
yards of yarn, the pounds per spyndle being: 

5. — 5 pounds. 

6. — 10 pounds. 

7. — 15 pounds. 

8. — 20 pounds. 

9. — 25 pounds. 
10. — 30 pounds. 

LESSON XXXIX.— Written. 

1. — Make out a table showing the weight in grains of 50 
yards of yarn for every pound difference in weight of a spyndle 
from 1 pound to 30 pounds. 



42 LUDLOW TEXTILE ARITHMETIC 

2. — Find circumference of rollers from 1 inch to 12 inches, 
for each inch difference. 

LESSON XL.— Written. 

1. — Show the weight of 50 yards of yarn of each lea from 
1 to 10 leas. 

2. — What would be the length of a bell with a 90 bell 
wheel, if the number of yards which passes under the drum 
of a lapper, with an 84 bell wheel, is computed to be 256 
■ yards ? 

(On certain machines, an attached bell rings automati- 
cally when a certain number of yards has been delivered. 
This length of material is called the length of bell.) 

3. — If a 46 draft wheel on a card is making a rove to 
weigh 6 ounces per 100 yards, what wheel would be required 
to produce a 7 ounce rove? 

4. — How many bales of jute would be required to make 
18,000 lbs. of carpet yarn, if 360 pounds of yarn is the aver- 
age yield of yarn from each 400 pound bale of jute? 

5.— What would be the cost of 100,000 lbs. of jute if 
2240 lbs. cost £26? 

(£1=$4.86) 

6. — With an 18" drum on the shaft, the spindles of a 
spinning frame are running at the rate of 2666 R. P. M. 
What would be the speed of these spindles with a 19" drum? 

7. — If a certain weight rove requires a draft of 6 on the 
spinning frame to make it into yarn weighing 14 lbs. per 
spyndle, what draft would be required to spin 10 lbs yarn 
from the same rove? 

8. — An 18" drum is driving the cylinder of a spinning 
frame at the rate of 400 R. P. M. and it is required to speed 
up this cylinder to 420 R. P. M. What diameter of drum 
will be required on the shaft to make this speed? 



FRACTIONS. 



Fractions occur in mill work and machine shop practice, 
in measuring diameters of rollers, shafts, and other parts of 
the machinery, but in nearly all other calculations, decimals 
are used. 

A fraction is a single part or a number of parts of a whole 
number. 



LUDLOW TEXTILE ARITHMETIC 43 

In the mill system of measurement, any dimension less 
than an inch is usually expressed in fractional parts of an 
inch, as 7/8", 9/16", 11/32", 5/64*. 

A fraction is expressed by two numbers, or terms, the 
numerator and the denominator. The denominator is 
written below the numerator with a line placed between. 

The denominator expresses the number of equal parts 
into which the unit is divided, and the numerator expresses 
the number of such parts denoted by the fraction. 

In the fraction 2/3, 2, is the numerator and 3, the de- 
nominator. 

A Proper Fraction is one whose numerator is less than the 
denominator, as 11/12, 4/7. 

An Improper Fraction is one whose numerator is not 
less than the denominator, as 13/6, 29/13. 

A Mixed Number is composed of both a whole number 
and a fraction, as 10 4/11, 2 1/3. 

If the terms of any fraction be both multiplied or both 
divided by the same number, the resultant fraction is equal 
to the original. 

Both terms of the fraction are multiplied by 2. 
3X2 6 3 6 



5X2 10 5 10 
If anything is divided into ten equal parts and six are 
taken one will have the same quantity as if the article were 
divided into five equal parts and three were taken. 

A fraction is multiplied by any number, either by multi- 
plying its numerator, or by dividing its denominator by 
that number. 

A fraction is divided by any number by dividing its 
numerator or by multiplying its denominator by that number. 

The same operation can be performed on fractions as on 
whole numbers, that is fractions can be added, subtracted, 
multiplied, or divided. 

Before performing such operations, it is first necessary 
to know how fractions may be modified without changing 
their values in order that the operations refered to, may be 
performed. This is called reduction of fractions. 



44 LUDLOW TEXTILE ARITHMETIC 

TO REDUCE A MIXED NUMBER TO AN IMPROPER 

FRACTION. 

Example. Change 4 2/5 to an improper fraction, or, find 
how many fifths are contained in the mixed number 4 2/5. 

As the denominator, 5, tells 

into how many parts each unit 

4X5 = 20 is to be divided, and there are 

20 + 2 = 22 4 units, there will be 20 fifths 

Therefore 4 2/5 = 22/5 in the units. Adding these 20 

to the other 2 fifths, shown by 
the numerator, the total value 
is 22 fifths, or 4 2/5 = 22/5. 
Rule. Multiply the whole number by the denominator 
of the fraction, add to the product the numerator of the 
fraction, and use this sum as the new numerator and the 
denominator of the fraction as the new denominator. 
LESSON XLL— Oral. 
Reduce the following to improper fractions: 

1.— 2 1/3 6.— 8 4/5 

2.-3 1/2 7 — 6 2/7 

3.-3 4/5 8.— 10 1/10 

4.-5 3/100 9.— 12 4/8 

5.-5 7/8 10.— 11 1/9 

LESSON XLII.— Written. 
Reduce the following to improper fractions: 

1.— 6 7/8 6.-23 9/14 

2.— 3 5/9 7.-44 21/22 

3.— 12 2/3 8.-66 2/3 

4.— 17 5/11 9.-87 1/2 

5.— 19 5/6 10.— 98 7/64 

Additional examples may be written and solved by the 
pupils. 

TO REDUCE AN IMPROPER FRACTION TO A WHOLE 
OR MIXED NUMBER. 
Example. Change 49/15 to mixed number. 

The denominator is 15. 
Therefore in each unit there 
15 ) 49 (3 will be fifteen equal parts. 

45 In 49 there will be as many 

— units as 15 is contained 

4 times in it, which is 3 times. 

Therefore 49/15 = 3 4/15 There are 4 parts or 4 fif- 

teenths left over. Conse- 
quently 49/15 = 3 4/15. 



LUDLOW TEXTILE ARITHMETIC 



45 



Rule. To change an improper fraction to a mixed 
number, divide the numerator by the denominator. The 
quotient will be the number of whole units, and the remainder 
the numerator, and the denominator the denominator of 
the fractional part. 



LESSON XLIIL— Oral. 

Reduce to a mixed number: 
1.— 5/4 
2.— 13/6 
* 3.— 17/3 
4.— 23/11 
5.— 48/13 
6.— 37/12 

LESSON XLIV.— -Written. 

Reduce to a mixed number: 

1.— 37/12 7.-789/36 

2.— 57/20 8.— 419/53 

3.— 75/15 9.-242/29 

4.— 66/22 10.— 563/47 

5.— 39/19 11.— 783/87 

6.— 126/23 12.— 655/655 



7.— 51/10 

8.-43/7 

9.-37/5 

10.— 39/9 

11.— 16/5 

12.— 45/8 



13.— 2031/47 
14.— 1069/100" 
15.— 769/43 
16.— 129/94 
17.— 81/27 
18.— 144/36 



TO REDUCE A WHOLE NUMBER TO A FRACTION 
HAVING ANY GIVEN DENOMINATOR. 

Example. Reduce 7 to a fraction having the denominator 
5. 

As each unit contains five 
7X5 = 35 equal parts, 7 units will con- 

Therefore 7 = 35/5 tain 7 times 5 or 35 parts and 

7 will equal 35/5. 
Rule. Multiply the whole number by the proposed 
denominator, the product will be the numerator and the 
proposed denominator, the denominator of the fraction. 

LESSON XLV.— Oral. 



1.— Red 


uce 9 to llths. 


2.— ' 


5 " 12ths. 


3.— ' 


11 " 8ths. 


4.— ' 


12 " 12ths. 


5.— ' 


15 " 4ths. 


6.— ' 


20 " 5ths. 



46 



LUDLOW TEXTILE ARITHMETIC 



7.— Reduce 16 to 3rds. 



8.— ' 


17 ' 


halfs. 


9.— ' 


24 ' 


3rds. 


10.— * 


21 ' 


6ths. 


11.— ' 


25 ' 


4ths. 


12.— ' 


26 ' 


5ths. 


13.— ' 


40 ' 


8ths. 


14.— ' 


42 ' 


halfs. 


15.— ' 


50 ' 


6ths. 



TO REDUCE A FRACTION TO ITS LOWEST TERMS. 
Example. Reduce 48/96 to its lowest terms. 

As dividing both terms 
of a fraction by the same 
number does not change its 



48/96 = 24/48 

24/48 = 12/24 

12/24= 6/12 

6/12= 3/6 

3/6 = 1/2 

Therefore 48/96= 1/2 



value, both the numerator 
and the denominator are 
divided by 2. This gives 
24/48, which is equal to 48/96. 
Dividing again by 2 gives 
12/24 which in turn is 
divided by 2 giving 6/12. 
Dividing by 2 and then by 3 gives 1/2, which is the value 
of 48/96 reduced to its lowest terms, as 1 and 2 are mutually 
prime. As 2X2X2X3 = 24, both terms could have been 
divided by 24, thus obtaining the value 1/2, directly. 

Rule. Divide both terms, or quotients obtained by 
such division, by the common factors until the two terms 
are mutually prime. 

LESSON XLVL— Oral. 
Reduce to lowest terms: 

1.— 8/32 7.— 40/100 

2.— 16/18 8.-27/36 

i 3.-24/28 9.— 30/32 

4.— 9/27 10.— 24/64 

5.— 5/25 11.— 33/44 

6.— 25/125 12.— 45/100 

LESSON XLVIL— Written. 
Reduce to lowest terms: 

1.— 108/180 6.— 81/243 

2.— 84/100 7.— 216/396 

3.— 39/156 8.— 45/540 

4.-844/972 9.— 429/1190 

5.-333/696 10.— 825/1000 



LUDLOW TEXTILE ARITHMETIC 47 

TO REDUCE A FRACTION TO HIGHER TERMS. 

Example. Reduce 4/11 to 55ths. 

In the first fraction the de- 
nominator shows that the whole 
11 ) 55 ( 5 unit was divided into eleven 

55 parts and the numerator shows 

— that 4 parts were taken. If 

4x5 = 20 the whole unit is to be divided 

Therefore 4/11 =20/55 into 55 parts, then each of the 11 

original parts will be divided 

into 5 parts, and then each part 

will be only 1/5 of the original size. Thus there will be 5 

times as many parts taken as when the whole unit was 

divided into llths, or 5x4 = 20, or 4/11=20/55. 

Rule. Divide the new denominator by the original de- 
nominator and multiply this quotient by the original num- 
erator for a new numerator. 

LESSON XLVIIL— Oral. 



1- 


-1/2 to 8ths 


7.- 


—9/10 to lOOths 


2- 
3.- 

4.- 
5.- 
6.- 


-2/3 to 12ths 
-3/5 to 30ths 
-2/7 to 28ths 
-3/20 to 60ths 
-7/15 to 45ths 


8.- 

9. 

10.- 

11.- 

12.- 


—3/16 to 48ths 
—8/11 to 99ths 
—4/9 to 36ths 
—5/17 to 51sts 
—6/13 to 39ths 




LESSON XLIX.- 


-Written. 


1.- 
2- 
3.- 

4.- 
5.- 
6.- 


-7/12 to 144ths 
-6/11 to 143rds 
-15/16 to 112ths 
-7/25 to 400ths 
-9/26 to 234ths 
-8/27 to 135ths 


7- 

8- 

9- 

10- 

11- 

12- 


-22/23 to 460ths 
-19/20 to 600ths 
-13/33 to 198ths 
-17/42 to 210ths 
-11/48 to 336ths 
-12/13 to 260ths 



TO MULTIPLY A FRACTION BY A WHOLE NUMBER 

Example. Multiply 3/32 by 10. 

In this case the numerator 3, in- 
dicates that 3 parts are taken. If 
3/32x10 = 30/32 10 times as many are to be taken, 

= 15/16 then there will be 30. Thus there 

will be thirty thirty-seconds or 
30/32. This may be reducedto 15/16. 



48 LUDLOW TEXTILE ARITHMETIC 

Rule. Multiply the numerator by the whole number 
and place as a new numerator over the original denominator. 
Reduce to lowest terms. 

LESSON L.— Oral. 

Multiply: 

1.— 3/7 by 2 7.-3/8 by 16 

2.— 9/16 by 4 8.— 5/16 by 8 

3.-2/3 by 6 9.-7/64 by 16 

4.— 7/15 by 5 10.— 3/17 by 4 

5.— 11/12 by 6 11.— 6/25 by 100 

6.-4/77 by 11 12.— 5/7 by 14 

LESSON LI.— Written. 

1.— 17/24 by 72 7.— 11/50 by 220 

2.— 11/12 by 243 8.— 7/9 by 468 

3.— 9/10 by 450 9.— 23/12 by 76 

4.— 7/16 by 512 10.— 11/15 by 90 

5.— 21/25 by 85 11.— 17/32 by 128 

6.— 7/17 by 340 12.— 6/7 by 434 

TO MULTIPLY A MIXED NUMBER BY A WHOLE 

NUMBER. 

Rule 1. Change the mixed number to an improper frac- 
tion and proceed as above. 

Rule 2. Multiply the whole numbers together, then 
multiply the fraction by the multiplier and add the two pro- 
ducts for the final product. 

TO MULTIPLY A FRACTION BY A FRACTION. 

Example. Multiply 3/4 by 5/7. 

To multiply 3/4 by 5/7 means 

to obtain 5/7 of 3/4. The de- 

3/4X5/7 = 3X5/4X7 nominator 4, indicates that the 

== 15/28 original unit was divided into 4 

equal parts. To obtain 1/7 of one 

of these parts means that each 

part must be divided into seven additional parts, each one 

of which would be 1/28 of the original unit. Thus 1/7 of 

1/4 would be 1/28, and, as there are three fourths, there will 

then be 3 times as much or 3X1/28 which equals 3/28. But 

five sevenths are taken so there will be 5 times this or 5 X 3/28 

which equals 15/28. Therefore 3/4 X 5/7 = 15/28. 

Rule. Multiply the numerators together for a new 
numerator and the denominators together for a new denomi- 
nator. Reduce if possible. 



LUDLOW TEXTILE ARITHMETIC 



49 



Note. Factors common to both a numerator and a 
denominator should be cancelled before multiplying. 



LESSON LII,— Oral. 



Multiply: 
1.— 1/2 by 2/3 
2.-3/5 by 3/7 
3.— 2/11 by 3/5 
4.-4/7 by 5/8 
5.— 3/11 by 4/7 
6.-8/4 by 2/9 



7.-3/4 by 7/15 

8.-3/5 by 6/7 

9.-4/5 by 10/11 

10.— 1/12 by 2/5 

11.— 9/10 by 5/6 

12.— 7/12 by 6/7 



LESSON LIIL— Written. 



Multiply: 
1.— 5/9 by 18/45 
2.— 16/75 by 25/4 
3.— 28/100 by 45/76 
4.— 13/27 by 29/35 
5.— 19/81 by 36/95 
6.-23/94 by 17/69 



7.— 7/100 by 25/49 

8.— 16/36 by 18/24 

9.— 49/50 by 10/21 
10.— 2/3 by 6/10 by 5/9 
11.— 13/14 by 21/36 by 6/26 
12.— 5/6 by 75/102 by 18/125 



TO MULTIPLY A FRACTION BY A MIXED NUMBER 

Rule. Change the mixed number to a fraction and 
proceed as above. 

RECIPROCALS. 

Definition. A reciprocal of any number is the quotient 
obtained by dividing unity by the number. 

Example. Find the reciprocal of 5. 
1-5-5 = 1/5 

Therefore 1/5 is the reciprocal of 5. 

As 1 -5-1/5 = 5 the reciprocal of 1/5 is 5. Therefore it may 
be stated that the reciprocal of any number, whole or frac- 
tional, is the number itself, inverted. 

LESSON LIV.— Oral. 



Find the reciprocal of: 

1.— 3 
2.— 6 
3.— 7 
4.— 13 

5.— 17 
6.-22 



7.-7/9 

8.— 5/11 

9.-6/23 

10.— 3 2/7 

11.— 2 3/13 

12.— 6 5/6 



50 



LUDLOW TEXTILE ARITHMETIC 



TO DIVIDE A FRACTION BY A WHOLE NUMBER. 



3/10+3 = 
p/iox 1/? = 1/10 



Example. 3/10-5-3. 

It is evident that the numerator, 
3, indicates the number of parts that 
are expressed, and that if this num- 
ber is divided by 3, that the quo- 
tient will be 1, or that 1/10 will be 
the full quotient. In other words 3/10 divided by 3 will give 
1/10. By inverting, or finding the reciprocal of the divisor, 
and multiplying the desired result may also be obtained. 

Explanation, showing the reciprocal relation between 
dividing and multiplying. 

Divide 16 by 4. 

16+4 = 4 or 16X1/4 = 4 

Rule. To divide a fraction by a whole number invert 
the whole number and multiply. The result will be the 
quotient desired. 



Divide: 
1.— 2/3 by 6 
2.-3/5 by 9 
3.— 7/10 by 7 

4.-4/7 by 8 



LESSON LV.— Oral. 

5.-5/6 by 10 
6.— 6/11 by 18 
7.— 4/13 by 26 
8.— 10/9 by 25 



9.— 7/15 by 14 
10.— 7/9 by 21 
11.— 8/9 by 16 
12.— 25/32 by 50 



2/3-5/7 = 
2/3x7/5 = 14/15 



TO DIVIDE A FRACTION BY A FRACTION. 

Example. Divide 2/3 by 5/7. 

As the divisor is less than 

unity, the quotient will be 

larger than the dividend. 

Dividing by 1/7 will evidently 

give a quotient 7 times as 

large as when the divisor is 

unity. Dividing by 5 is equivalent to multiplying by 1/5. 

Thus, dividing by a fraction or multiplying by the reciprocal 

of the fraction, will produce the same result. 

Rule. Multiply the dividend by the reciprocal of the 
divisor. 

LESSON LVL— Written. 

Find the quotient of 
1.— 18/25 + 3/5 
2.— 5/6 + 18/25 



LUDLOW TEXTILE ARITHMETIC 



51 



3.— 4/9 -4- 7/36 

4.— 3/5-5-5/3 

5.— 17/28-5-15/34 

6.— 23/24-9/16 

7.— 11/25-5-14/15 

8—10/42-5- 5/21 

9.— 11/18-5-22/63 
10.— 6/7-5-7/6 
11.— 39/100-4-10/13 
12—46/47-5-23/48 
13.— 29/30-5-5/13 
14—26/43-5-52/86 

TO FIND WHAT PART ONE NUMBER IS OF 
ANOTHER. 

Example. 



4 divided by 12 = 
4/12 = 1/3 



What part of 12 is 4? 

1 is 1/12 of 12. Conse- 
quently 4 would be four 
times as much or 4/12 of 12. 
This may be reduced to 1/3. 
Therefore 4 is 4/12 or 1/3 of 12. 
Rule. Divide the number which is the part by the 
number of which the part is to be found. 

LESSON LVIL— Written. 



What part of 


1.— 25 


LS 5? 


2.-36 i 


LS 6? 


3.— 50 


LS 10? 


4.-48 


is 8? 


5.-33 ] 


s 11? 


6.-66 ] 


is 12? 


7.— $2 ] 


ls 50^? 


8.— $1 i 


is 16 2/3^? 


9.— $1 i 


ls 12 1/2^? 


10.— $1 


ls 37 1/2jzT? 


11.— $1 is 6 1/4^? 


12.— $10 


is 25^? 



TO CHANGE FRACTIONS TO SIMILAR FRACTIONS. 

Similar Fractions are fractions having a common 
denominator. 



52 LUDLOW TEXTILE ARITHMETIC 

Example. Change 3/5 and 1/2 to similar fractions. 
3/5 = 6/10 
1/2 = 5/10 

6/10 and 5/10 are similar 
Rule. Find the least common multiple of the denom- 
inators of the fractions for a common denominator, then 
find how many times each denominator goes into this L. C. M. 
'and, as the quotient is found, in each case, multiply it by 
the numerator of the fraction for a new numerator. 

LESSON LVIII.— Oral. 



Change to similar fractions: 




1.— 1/2, 1/3 


6.— 5/16, 5/8 


2.— 1/4, 1/3 


7.-2/7, 4/9 


3.— 1/5, 1/8 


8.-3/5, 7/15 


4.-2/3, 5/6 


9.— 3/10, 2/3 


5.-7/8, 3/4 


10.— 5/6, 7/8 


LESSON LIX.- 


—Written. 


Change to similar fractions: 




1.— 1/8, 3/4, 5/16 




2.-7/9, 5/6, 14/27 




3.-4/5, 9/50, 23/25 




4.— 9/10, 4/60, 7/12 




5.— 1/3, 1/2, 2/5 




6.-7/23, 1/2, 9/46 




7.— 14/35, 9/14, 3/7 




8.— 7/15, 1/3, 5/12 




9.-9/24, 2/3, 3/16 




10.— 5/18, 4/5, 7/9 


• 



ADDITION OF FRACTIONS OR MIXED NUMBERS. 

Example. Add 3/4, 5/8, 7/16. 

Change all the fractions 

to similar fractions. The 

3/4 — 12/16 three fractions will then 

5/8 =10/16 become 12/16, 10/16 and 

7/16= 7/16 7/16, that is, there are 12 

parts in the first, 10 in the 

29/16=1 13/16 second, and 7 in the third 
or 29 parts in all, or 29/16. 
This may # be reduced to 
the mixed number 1 13/16. 



LUDLOW TEXTILE ARITHMETIC 53 

Rule. Change to similar fractions, if necessary, and 
add numerators together for a new numerator and place over 
the common denominator. 

Example. Add 2 3/4, 5 5/8, 4 7/16. 

The fractional parts 

are changed to similar 

2 3/4 =2 12/16 fractions, added to- 

5 5/8 =5 10/16 gether, and reduced if 

4 7/16 = 4 7/16 possible. The whole 

numbers are added to- 



ll 29/16 = 12 13/16 gether and this sum is 
added to the sum of the 
fractional parts. 
Rule. Add the fractional parts together and then the 
whole numbers and find the sum of the two results. 

LESSON LX.— Oral. 

Add: 

1.— 1/8, 1/4 6.— 1/2, 7/16 

2.— 3/16, 5/8 7.-3/4, 7/8 

3.— 5/16, 3/8 8.-3/64, 3/8 

4.— 1/2, 3/8 9.-5/32, 5/64 

. 5.— 7/16, 1/8 10.— 7/8, 9/64 

LESSON LXL— Written. 

Add: 

1.— 15/16, 7/8, 3/4 

2.— 3/10, 1/5, 6/25 

3.— 9/21, 4/7, 1/5 

4.— 7/12, 5/6, 7/8 

5.-25 1/4, 46 7/16 

6.-36 1/8, 49 7/16, 15 7/12 

7.— 13 4/15, 12 7/18, 9 1/6 

8.-5/7, 1/4, 5/6, 19/21 

9.— 11 7/64, 9 3/4, 14 13/32, 16 1/2 
10.— 29/36, 5 1/6, 14 7/9, 2/3 

SUBTRACTION OP FRACTIONS OR MIXED NUMBERS. 

Example. Subtract 3/5 from 7/8. 

First change the fractions to 

similar fractions. *7/8 becomes 35/40 

7/8 = 35/40 and 3/5 24/40, that is, in the former 

3/5 = 24/40 there are 35 parts and in the latter 

there are 24 parts. The difference 

11/40 is 11 parts or 11/40. 



54 LUDLOW TEXTILE ARITHMETIC 

1 "Example. Subtract 6 3/4 from 12 3/10. 

Change the fractional 

12 3/10 = 12 12/40 = 11 52/40 parts to similar fractions. 

6 3/4 = 6 30/40= 6 30/40 As 30/40 is greater than 

12/40 the former cannot 

5 22/40 be subtracted from the 
latter. Take one whole 
5 22/40 = 5 11/20 one from the 12 leav- 

ing 11 whole ones and 
change this whole one to fortieths. Add these 40 parts to 
the 12 thus making 52 parts or 52/40. Then the 30 parts can 
be subtracted from the 52 parts leaving 22 parts or 22/40 
and the 6 whole ones can be subtracted from the 11 leaving 
5. As 22/40 may be reduced to 11/20 the whole remainder 
becomes 5 11/20. 

Rule. Change fractional parts to similar fractions; if 
the numerator of the minuend is less than that of the subtra- 
hend, borrow one unit from the whole number of the minuend, 
change it to a similar fraction and combine with the frac- 
tional minuend. Subtract the whole numbers. Then sub- 
tract the numerator of the subtrahend from the numerator 
of the minuend and place the result over the common de- 
nominator as the fractional part of the remainder; combine 
the two remainders for total difference. 

LESSON LXIL— Oral. 

Find value of: 
1.— 1/2— 1/4 6.— 15/16— 5/8 

2.— 7/10— 3/10 7.-5—3/4 

3.— 7/8— 1/2 8.— 22— 9/16 

4.— 9/16— 3/8 9.-27—5/8 

5.— 11/12— 5/8 10.— 36— 11/64 

LESSON LXIIL— Written. 

Find the value of: 
1.— 27— 3 5/8 
2.-29—6 3/4 
3.— 42— 21 7/16 
4.-28 1/2—13 3/8 
5.-45 3/4—11 15/16 
6.— 12 7/8—2 63/64 
7.— 18 3/64—17 5/8 
8.-6 47/100—4 2/5 
■ 9.— 10 3/8—6 3/5 

10.— 1/2 + 1/3—1/4 + 1/5—1/6 



LUDLOW TEXTILE ARITHMETIC 55 

TO CHANGE A DECIMAL TO A COMMON FRACTION. 

Example. Change 0.06 to. a common fraction. 

As the zero at the left of the 6 

0.06 = 06/100 = 3/50 does not affect the value of num- 
ber expressed as numerator, it 
may be omitted. By cancellation 6/100 is reduced to its 
lowest terms 3/50. 

Rule. For a numerator take the figure or figures ex- 
pressing the decimal and for a denominator take the figure 
one with as many zeros as there are figures in the decimal, 
counting each zero as a figure. 

LESSON LXIV.— Written. 

Change to common fraction: 

1.— 0.2 11.— 3.96 

2.— 0.6 12.— 6.666 

3.— 0.50 . 13.— 7.125 

4.— 0.25 14.— 4.0625 

5.— 0.75 15.— 12.096 

6.— 0.305 16.— 0.0005 

7.— 0.06 17.— 14.009 

8.— 0.006 18.— 7.375 

9.— 0.3245 19.— 0.00025 

10.— 5.4008 20.— 10.4238 

MISCELLANEOUS EXAMPLES. 

LESSON LXV.— Oral. 

1. — A certain grade of carpet is made from 1/4 flax, 
1/3 hemp, and the remainder jute. What part of the carpet 
is jute? 

2. — What is the cost of 80 pounds of flax at 12 1/2 
cents a pound? 

3. — What is the cost of a 400 pound bale of jute at 5- 
3/4^ a pound? 

4. — What is the difference in the price of 400 pounds of 
jute at 6 3/4^, and the same weight of jute at 6 15/16^ per 
pound? 

5.— What are the wages per day of 10 hours of a boy 
who is paid at the rate of 9 1/4 cents per hour? 

6. — If a box will hold 1 3/4 bales of jute, how many 
bales will it take to fill 12 boxes? 



56 LUDLOW TEXTILE ARITHMETIC 

7. — If a girl is paid 3 3/4^ for balling 6 pounds of twine, 
how much will she receive for balling one pound? 

8.— If one card uses, on an average, 7 3/4 bales of jute 
in a day, how many bales will 11 cards use in a day? 

9. — A roving frame runs 10 hours a day. How many 
doffs will be taken from it if it produces an average of 2 3/4 
doffs per hour? 

10. — What would it cost to pack 25 barrels of twine at 
2 3/8 cents a barrel? 

11. — How many quarter pound balls of twine can be 
made from 11 1/2 lbs. of twine? 

12. — How many 1/2 pound balls of twine are contained 
in a barrel weighing 179 pounds, making no allowance for 
weight of barrel? 

LESSON LXVI.— Written. 

1. — A bale of flax costs $33.84, which is at the rate of 
9 3/4 cents a pound. How many pounds are contained in 
the bale? 

2. — Find the cost of spinning 105,000 pounds of yarn 
at 7/8^ per pound. 

3. — What is the cost of manufacturing 1000 lbs. of twine, 
if preparing and spinning cost 7/8^ per pound, and twisting 
and finishing cost 15/16^ a pound? 

4. — What is the cost of a bale of flax containing 224 
pounds at 16 2/3^ a pound? 

5. — At 8 3/5 $ a pound, how many pounds of jute can 
be bought for $1000? 

6. — If castings cost 1 1/5/ a pound, what part of a ton 
can be bought for $14 1/2? 

7.— What is the cost of rove bobbins per gross if a single 
bobbin costs 5 7/10^? 

( 1 gross = 144 ) 

8. — What is the cost of 64 sets of roving spindles and 
flyers, if each set costs $1 3/4? 

9. — -What is the cost of 10 gross of screws if each dozen 
costs 15 2/10^? 

10. — How many yards of 6 lea yarn is contained in 1/300 
part of a lb. ? 

11. — What is the difference in the circumference of two 
rollers, one of which measures 2 1/4 inches, and the other 
2 3/16 inches? 



LUDLOW TEXTILE ARITHMETIC 57 

12.— What is the cost of 7 1/4 lbs. of flax at 10 1/2 jzC a 
pound? 

13.— What is the cost of 6 2/3 lbs. of tow at 3 3/4$^ a 
pound. 

14. — What is the cost of 15 1/2 lbs. of jute at 5 1/4^ a 
pound? 

15.— What is the cost of 20 7/8 lbs. of brass at 26 1/2^ 
a pound? 

16. — -Find the cost of 1000 lbs. of a mixture, 1/4 of which 
cost 5 l/2jzf a pound, 1/3 of which cost 4 3/4^ a pound, and 
the remainder at the rate of 6 cents a pound. 

17. — How many balls, each containing 1/6 of a pound, 
can be made from 17 2/3 pounds of twine? 

18. — A certain grade of yarn is made from stock, 1/2 of 
which costs 72/8^ per pound, 1/4 at 67/8$/ per pound, and 
1/4 at 60/8^ per pound. What is the average cost per pound 
of the stock used? 

19. — -What does a boy receive for a week's wages of 55 
hours, who is paid at the rate of 10 9/10^ an hour? 



WEIGHTS AND MEASURES. 

Weighing is the first operation that is performed on every 
bale of raw stock which comes to Ludlow, and weighing is 
the last thing that is done to every package of manufactured 
goods which leaves the town. Moreover the fibre is weighed 
many times in the course of its many stages of manufacture. 

It is measured and weighed in Laps before it goes to the 
Finisher Cards ; and the Rove is measured and weighed. 
The fibre is measured and weighed again as Yarn and as 
finished Ropes and Twines. All of the calculations about 
the different machinery are worked out from the weights 
and measurements, and the goods are finally marketed to 
equal a certain standard measurement for a certain standard 
weight. 

Avoirdupois Weight is the standard by which all the 
goods in these mills are bought and sold and it is also the 
standard upon which all calculations in this work are based. 

The Unit of Measurement for goods in process of 
manufacture and for the manufactured product is the number 
of yards contained in one pound avoirdupois, the latter being 
equal to 7000 grains. 



58 LUDLOW TEXTILE ARITHMETIC 

All the raw stock arrives in Ludlow in bales. 

Jute bales weigh 400 pounds each (average). 

Flax and tow weigh from 200 to 600 pounds each bale. 

Hemp of different kinds weighs from 400 to 600 pounds 
each bale. 

Manufactured goods are made up for shipment in Balls, 
Coils, Reels, Bags, Boxes, Barrels, and Bales. In the' 
bagging and webbing departments, the finished cloth is made 
to average a certain weight per Roll. 

In the preparing departments the weighing and the 
measuring of the material begins after it has passed through 
the breaker cards. The sliver is run over a Lapping ma- 
chine, to the calendar roller of which is attached the mech- 
anism which records the number of yards contained in each 
lap. When a certain number of yards of sliver have been 
rolled together a bell attached to the machine, rings. The 
lap is then taken off and weighed. Three of these laps com- 
bined, or whose combined weights total a certain number 
of pounds (usually 250 lbs.), are formed into what is known 
as a Sett. These setts of laps are then passed through the 
finisher cards, and upon their uniformity depends the uni- 
form weight of the yarn which is to be made from the material. 

On the weight and length of these setts of laps are based 
the calculations for the Drafts of the card, drawing, and 
roving frames, to produce a rove which will measure a cer- 
tain number of yards per ounce, or, what amounts to the same 
thing, will weigh a certain number of ounces to the 100 yards. 

The rove is a slightly twisted sliver and the weight of 
this rove regulates the size of the Draft Wheel which is 
put on the spinning frame to produce the yarn which will 
weigh a given number of pounds per spyndle. 

In the preparing room the rove is weighed daily, usually 
by measuring 100 yards from several bobbins and then tak- 
ing the average weight as the standard for the calculations. 

The Draft to be put on the spinning frame is found by 
dividing the weight of the yarn required into the weight of 
the rove. 

After the yarn is spun the drawing out process ceases. If 
the yarn is destined for twines or ropes, the different weights 
are obtained by additional strands. 

The calculations for finding the length of bells and the 
drafts on different machines will be found in the chapter on 
general mill work. 



LUDLOW TEXTILE ARITHMETIC 59 

LUDLOW YARN TABLE. 

1 thread = circumference of a 90" reel — 2 1/2 yards 

120 threads = 1 lea or cut = 300 yards 

48 cuts = 1 spyndle — 14,400 yards 
The Lea of the yarn is designated by the number of leas 
or cuts in lib., that is, the number of times that 300 yards is 
contained in one pound avoirdupois. For example, if one 
pound of yarn contains 1800 yards, divide the 1800 by 300 
and the quotient will be the lea. 

1800-=- 300 = 6 lea 

(This would be read sixes or 6 lea and would be written 
6's.) 

Pounds Per Spyndle. The pounds per spyndle are the 
number of pounds contained in 14,400 yards. For example, 
if 14,400 yards of yarn weigh 14 pounds the yarn is called 
or named 14 lbs. yarn. 

(Spyndle is spelled in this manner by most authorities on 
jute spinning to distinguish the word from the- word spindle, 
although there is no authority for this method in the Stand- 
ard Dictionary.) 

The weight of 3 cuts or 900 yards, in ounces, is equal to 
the pounds per spyndle. 

3 cuts = 900 yards 

1 spyndle = 14,400 yards 

.". 3 cuts or 900 yards -f- 14,400 yards = 1/16 of a spyndle. 

As 16 ozs. = 1 pound there would be as many ounces in 3 
cuts as there would be pounds in one spyndle. 

To reduce leas to pounds or to find the pounds per spyndle 
having the lea given. 

Rule. Divide the number of the lea into 48 and the 
result is pounds per spyndle. 

If the yarn is 6 lea then: 

48-^6 = 8 or 8 lbs. yarn. 

To change pounds per spyndle to leas, reverse the rule 
above, that is divide 48 by pounds per spyndle and the 
result is leas per pound. For example, for 14 lbs. yarn. 

48 -5- 14 = 3 . 43 leas per pound. 

Reeled Yarn is designated by the leas per pound. 

Yarn shipped on Quills is designated by pounds per 
spyndle. 



60 LUDLOW TEXTILE ARITHMETIC 

LUDLOW YARN TABLE EXAMPLES. 

LESSON LXVL— Written. 

1. — Find the weight in grains, of 100 yards of yarn 
which weigh 10 lbs. to the spyndle. 

. 2. — If a spyndle of yarn is equal to 12 lbs. what is the 
lea of the yarn? 

3. — How many threads are there in 100 yards of 14 lbs. 
jute yarn? 

4. — What is the weight, in grains, of 50 yards of 4 lea 
yarn? 

5. — If 50 yards of yarn are equal to 243 grains, what are 
the pounds per spyndle of that yarn ? 

6. — Find the lea and pounds per spyndle of yarn which 
weighs 467 grains to the 60 yards. 

7. — What is the weight of 100 yards of yarn which 
weighs 24 lbs. per spyndle? 

8. — What are the pounds per spyndle of yarn which 
weighs 729 grains per 50 yards? 

9. — Find the leas per pound when 7500 yards of yarn 
equal one pound. 

10. — What would be the weight, in grains, of 300 yards 
of 25s or 25 lea yarn? 

11. — How many yards are there in one ounce of yarn 
which weighs 14 lbs. per spyndle? 

12. — Find the number of yards contained in three ounces 
of 8 lea yarn. 

13. — Find the weight in ounces of 900 yards of 6 lbs. 
jute carpet yarn. 

14. — How many spyndles of 14 lbs. yarn are contained 
in 84,000 pounds? 

15. — Find how many cuts of 6 lbs. carpet yarn is contained 
in a bale whose net weight is 640 lbs. 

16. — Find how many cuts of 8 lea carpet yarn there are 
in a bale the same weight as in example 15. 

17. — How many spyndles of 15 lbs. carpet yarn would 
be required to fill an order for 120,000 pounds of this particu- 
lar size yarn? 

18. — Deducting 1/10 for loss in manufacture, how many 
spyndles of 10 lbs. yarn would be produced from a bale of 
jute weighing 400 lbs. ? 

19. — Find the weight, in grains, of 100 yards of yarn for 
every pound difference in weight of spyndle from one pound 
to 10 pounds, both inclusive. 



LUDLOW TEXTILE ARITHMETIC 61 

20. — Make out a table showing the weight, in grains, of 
50 yards each different lea of yarns from l's to 10's, both 
inclusive. 

NUMBERING FOR TWINES AND ROPES. 

The standard of measurement for fine twines is the num- 
ber of yards contained in one pound. 

Twines up to 36 size are 3 ply. 

to 48 size are 4 ply. 

- to 60 size are 5 ply. 

to 72 size are 6 ply. 

Balings and jute wrapping twines are twisted 2, 3, 4, 5, 6, 
7 and 8 ply. 

Seaming and other special twines are made to order. 

Binder Twines. The size of binder twines is indicated 
by the number of feet per pound. There are four standard 
sizes, namely, 500, 550, 600, and 650 feet per pound. 

Rope Yarns. The size or number of rope yarn indicates 
the number of threads of that yarn which will be required 
to make one of three strands which will form a rope 3 inches 
in circumference. For example, No. 30 indicates that 3 
strands of 30 threads each, or 90 threads in all, will make a 
rope 3 inches in circumference. The usual sizes range from 
12's to 40's. 

The weight of 100 yards of No. 40 may be calculated as 
follows: 

The weight of 100 yards, 3 inches in circumference, white 
rope, averages about 84 pounds. The contraction by twist 
being about 25 per cent., {%), each of the single strands of 
yarn composing the rope must have a length of 125 yards, or 
the total length of the 120 strands will be 15,000 yards. 
Since the length equals 84 lbs. which is equal to 1344 ozs., 
100 yards weigh nearly 9 ozs. 

Similarly: 

No. 20 rope yarn — 18 ozs. per 100 yards. 
No. 30 rope yarn = 12 ozs. per 100 yards. 
No. 18 rope yarn = 20 ozs. per 100 yards. 

LONG MEASURE. 

12 inches — 1 foot 

3 feet = 1 yard 

1760 yards or 5280 feet = l mile 

h}/2 yards or 163^ feet — 1 rod. 



62 LUDLOW TEXTILE ARITHMETIC 

REDUCTION OF COMPOUND QUANTITIES, 
DESCENDING. 

Reduce 3 yds. 2 ft., 6 in. to inches, 
yds. ft. in. 
3 2 6 

3 

9 

2 As each yard contains 3 ft. 

— change yards to feet by multiply- 

11 ft. ing by 3, and then add the two 

12 feet to this product. Change the 
1 1 feet to inches by multip lying by 

22 12, add the 6 inches, making a 

11 total of 138 inches. 

132 



138 in. 
Rule. Multiply the quantity of largest denomination 
by the number which shows how many of the next lower 
denomination are required to make one of the higher denomi- 
nation and add the units of lower denominations to this 
product. Repeat this process until the quantity of lowest 
denomination is reached. 

REDUCTION OF COMPOUND QUANTITIES, 
ASCENDING. 

Example. Reduce 197 inches to higher denominations. 

As 12 inches equal 1 ft., 

dividing 197 by 12 will give the 

12 ) 197 inches value in feet which is 16, with 

5 in. over. As 3 feet equal 1 

3 ) 16 feet, 5 inches yard, dividing by 3 will give 

5 yards, 1 foot the number of yards. 16-^3 

5 yds., 1 ft., 5 in. = 5 yds. and 1 ft. over. Thus 

197 inches equals 5 yds., 1 ft., 
5 in. 
Rule. Divide by the number which represents the num- 
ber of units necessary to equal one of the next higher denomi- 
nation. The quotient will be the number of units of the 



LUDLOW TEXTILE ARITHMETIC 63 

higher denomination and the remainder will be the number 
of the lower denomination. Repeat the same process with 
the quotient until it is impossible to reduce to a higher denomi- 
nation. The final value will consist of the sum of the last 
quotient and all the remainders arranged in order. 

LESSON LXVIIL— Written. 

1. — Reduce 6 yds., 1 ft., 3 in. to inches. 

2. — Reduce 4 yds., 2 ft., 11 in. to inches. 

3. — Reduce 432 inches to yds. 

4.-397 inches are to be reduced to yds. 

5. — 16,492 ft. are to be reduced to higher denominations. 

6. — Reduce 33,764 feet to higher denominations. 

7. — Reduce 7 yds., 1 ft., 6 in. to inches. 

ADDITION AND SUBTRACTION OF COMPOUND 
QUANTITIES. 

Example. 'Add 3 yds., 2 ft., 11 in., to 1 yd., 2 ft., 7 in. 
Write, in a row, the names of the 
quantities of different denominations, 
yds. ft. in. and then place the different quanti- 

3 2 11 ties in their proper columns. Add the 
12 7 figures in the separate columns, com- 
mencing with the units of lowest de- 

4 4 18 nomination at the right. 11" + 7" = 18" 

-1' 6". Add the 1' to the next col- 

5 2 6 umn making 5' which = 1 yd., 2 ft. 

Carry the 1 yd. to the next column 
and add to the 3 and 1 making 5 yds. 
The total will then be 5 yds., 2ft., 6 in. 

(Always arrange quantities of different denomination in 
regular order, descending from left to right.) 

Example. Subtract 3 yds., 1 ft., 6 in. from 15 yds., 1 ft., 
5 in. 

Arrange the quantities of different 
denominations in separate columns. 
6 in. from 5 can not be taken so 1 ft. is 
borrowed from the next column. 1 ft. 
equals 12 in. 12 in. +5 in. equals 17 
in. 17 in. — 6 in. = 11 in. 1 ft. from 
11 2 11 ft. can not be taken. Borrow 1 yd. 

which equals 3 ft. 3 ft. minus 1 ft. 
= 2 ft. 3 yds. from 14 yards equals 
11 yds. Thus the remainder will be 11 yds., 2 ft., 11 in. 



yds. 


ft. 


m. 


15 


1 


5 


3 


1 


6 



64 



LUDLOW TEXTILE ARITHMETIC 





LESSON LXIX.— Written. 




Add: 


1. 






2. 


* 


yds. 


ft. 


in. 


yds. 


ft. 


in. 


6 


1 


3 


13 


1 


10 


2 





11 


15 


2 


8 


3 


2 


8 


19 


2 


9 




3. 






4. 




mi. 


yds. 


ft. 


mi. 


yds. 


ft. 


7 


1240 


2 


3 


1648 


1 


12 


698 


2 


7 


1339 


2 


3 


1443 


2 


8 


1547 


2 


Subtract : 


5. 






6. 




yds. 


ft. 


in. 


yds. 


ft. 


in. 


33 


1 


7 


29 


2 


7 


21 


2 


9 


20 


2 


11 




7. 






8. 




mi. 


yds. 


ft. 


mi. 


yds. 


in. 


14 


1146 


1 


10 


1496 


2 


11 


1539 


2 


3 


1500 


1 


fLTIPLIC 


IATIO* 


r AND DIVISION OF 


COMPOUND 






QUANTITIES. 






Example 


. Multiply 7 yds., 


, 2 ft., 2 : 


in. by 6 





yds. 

7 



ft. in. 6 times 2 in. = 12in. = 1 ft. 

2 2 6 times 2 ft. = 12 ft., 12 ft. + l 

6 ft. = 13 ft. =4 yds., 1 ft. 6 

times 7 yds. =42 yds. 42 yds. 

+ 4 yds. =46 yds. Total value 

equals 46 yds., 1 ft., in. 

Divide 432 yds., 1 ft., 9 in. by 3. . 

3 goes into 432 yds. 143 with 
a remainder of 2 yds. 2 yds. = 6 
ft. 6 ft. added to 1 ft. =7 ft. 3 
goes into 7 twice with a remainder 
of lft. 1ft. = 12 in. 12 in. +9 in. = 
143 2 7 21 in. 3 goes into 21 7 times. 

Final value equals 143 yds., 2 ft., 
7 in. 



46 




13 


12 


46 




1 





Example. 


Di 


vide 


3 


yds. 
) 432 


ft. 

1 


in. 
9 



LUDLOW TEXTILE ARITHMETIC 65 

LESSON LXX.— Written. 

1.— Multiply 15 yds., 1 ft., 6 in. by 12. 

2.— Multiply 13 yds., 2 ft., 8. in. by 23. 

3.— Multiply 2 mi., 1120 yds., 2 ft. by 7. 

4.— Multiply 6 mi., 1394 yds., 1 ft. by 9. 

5.— Divide 176 yds., 2 ft.", 7 in. by 8. 

6.— Divide 238 yds., 1 ft., 9 in. by 12. 

Note. All compound quantities may be manipulated in 
the same way as the previous operations were performed in 
long measure. 

LESSON LXXI.— Written. 

1. — -How many feet of belting would be required to dri~ 7 e 
96 frames if it takes 24 ft., 6 in. for one frame? 

2. — How many yards of tape would be required to drive 
the spindles of 6 dry spinning frames each frame having 128 
spindles, and the length of tape for each spindle being 68 
inches ? 

3. — How many 28 feet length rails would be required to 
lay a double track railroad from Ludlow Bridge to Palmer, a 
distance of about 8 miles? 

4. — How many yards from Ludlow to New York, a dis- 
tance of 145 miles? 

5. — How many miles of thread would be contained in 
60,000 pounds of yarn weighing 5 lbs. to the spyndle? 

6. — A spyndle of yarn contains 14,400 yards. How 
many spyndles would be required to reach from Ludlow to 
Boston, a distance of about 90 miles? 

7. — *How many 22 feet lengths of shafting would be re- 
quired to put 3 lines of driving shafts in each of the four 
floors of No. 16 Mill? The length of each room being 19S feet. 

8. — If the net weight of rove on a bobbin is 2}/% pounds, 
how many yards does it contain if each 100 yards of material 
weighs 8 ounces ? 

SQUARE MEASURE. 

Table. 
144 square inches = 1 square foot. 
9 square feet = 1 square yard. 
30% sq. yards or 272 1/4 sq. feet = 1 square rod. 
160 square rods or 4840 square yards or 43560 square feet- 
= 1 acre. 

640 acres = 1 square mile. 

An acre equals a square whose side is 208.71 feet. 



66 LUDLOW TEXTILE ARITHMETIC 

LESSON LXXIL— Written. 

1. — Find the total floor space in the four floors of a Flax 
Mill, the dimensions of each floor being 300 ft. X66 ft. 

2. — How many square feet of floor space are contained 
in one of the floors of a mill building, the inside dimensions 
being 198 ft. by 88 ft.? ' 

3. — Find the floor space in square yards contained in the 
five floors of a Mill, the inside dimensions being 176 ft. 
X66 ft. 

4. — It is estimated that there are 4^ acres of floor space 
in two mills. How many square feet of floor space would 
that equal? 

5. — How many yards of plastering are required to cover 
four sides of a room, 18 ft. long, 15 ft. wide, and 9 ft. high, 
making no allowances for doors or windows? 

6. — How many yards of carpeting a yard wide will it take 
to cover a floor 18 feet square? 

7. — How many square yards of cement sidewalk are con- 
tained in a rectangle two sides of which are 1875 feet and the 
other two sides are 520 feet, the walk being 43^ feet wide 
and extending on the four outside edges of the rectangle? 

8. — How many square yards of carpeting will it take to 
cover a room 4 yards long and 3 yards wide? 

9. — How many square yards in a ceiling 35 ft. long and 
28 ft. wide? 



CUBIC MEASURE. 

TABLE.' 

1728 cubic inches = 1 cubic foot. 
27 cubic feet = 1 cubic yard. 
128 cubic feet = 1 cord. 
7 . 48 gallons = 1 cubic foot. 

LESSON LXXIIL— Written. 

1. — How many cubic feet in a tank 8 feet long, 4 feet 
wide, and 3 feet deep? 

2. — How many cubic feet in a load of wood 8 feet long, 
43^ feet high and 3^ feet wide? 

3. — How many cubic feet in a cistern 15 feet long, 12 
feet wide, and 10 feet deep? 



LUDLOW TEXTILE ARITHMETIC 



67 



4. — What is the cubic capacity of a room in No. 12 Mill, 
the inside dimensions of which are 198 feet long, 88 feet wide, 
and 16 feet high? 

5. — How many cubic feet in a tank which holds 6000 
gallons of oil? 

6. — How many cubic feet in a tank which holds 50 
barrels of oil? 

(31^gal. = lbbl.) 

7. — What would be the number of cubic feet of space in 
a room 176 feet long, 66 feet wide, and 16 feet high? 

8.— How many bales 4 feet long, 3 feet wide, and 2 feet 
thick could be stored in a stock house 100 feet X 100 feet, 
and 16 feet high making no allowances for waste spaces? 

9. — How many cubic yards in a cellar 18 feet long, 12 
feet wide, and 9 feet deep? 

10. — What are the cubic contents, in inches, of a bale 
measuring 4 ft. by 3 ft. by 2 ft. ? 



LIQUID MEASURE. 



TABLE. 



4 gills = 1 


pint. 










2 pints = 


1 quart. 








4 quarts = 


= 1 gallon. 








31/^2 gallons = 


1 barrel. 








2 barrels 


= 1 hogshead. 










LESSON LXXIV.— Written. 




Add: 


1. 






2. 




gals. 


qts. 


pts. 


gals. 


qts. 


pts. 


25 


3 


1 


21 


3 





7 


1 





16 


3 


1 


5 


2 


1 


22 


3 


1 


6 


3 


1 


5 


3 


1 


Find the difference between : 










3. 






4. 




gals. 


qts. 


pts. 


gals. 


qts. 


pts. 


10 


2 





25 


1 





4 


2 


1 


16 


2 


1 



68 LUDLOW TEXTILE ARITHMETIC 

5. — How many pints of oil are contained in a tank of 
5760 gallons? 

6. — After 2044 gals, and 3 qts. have been taken from a 
tank containing 5760 gallons, how much remains? 

7.— Multiply 8 gals., 2 qts., 1 pt. by 8. 

S — Multiply 15 gals., 1 qt., 1 pt. by 7. 

9.— Divide 116 gals., 2 qts. by 4. 
10.— Divide 162 gals., 3 qts. by 6. 



AVOIRDUPOIS OR COMMERCIAL WEIGHT. 

TABLE. 

437 . 5 grains = 1 ounce. 
70Q0 grains = 1 pound. 

or 
16 ounces = 1 pound, 
100 pounds = 1 hundredweight. 
2000 pounds = 1 ton. 
112 pounds = 1 long hundredweight. 
2240 pounds = 1 long ton (British). 

LESSON LXXV«— Written. 

1. — How many grains in 3 ounces of yarn? 

2. — Reduce 37 tons, 19 hundredweight, 3 grains to 
p junds. 

3. — Reduce 1,742,684 grains to pounds? 

4. — How many grains are contained in a spyndle of 
yarn weighing 14 lbs. ? 

5. — Reduce 574,692 ozs. to units of higher denomination. 

6.— How many tons in 100,000,000 lbs.? 

7. — In a cargo of jute containing 6400 bales, there are 
h >w many tons? 

8. — What would be the weight of 10,000 pins if 78 pins 
weigh one ounce? 

9. — What would be the length of a roll of wire weighing 
50 lbs., if the weight of one foot was 87^ grains? 

10. — What is the cost per pound of American hemp that 
costs $150 per ton? 



LUDLOW TEXTILE ARITHMETIC (59 

ADDITIONAL TABLES. 

DRY MEASURE. 



2 pints = 1 quart. 
8 quarts = 1 peck. 
4 pecks = 1 bushel. 

SHIPPING MEASURE. 

40 cubic feet = 1 U. S. shipping ton 
42 cubic feet = 1 British shipping ton. 

NUMBERS. 

12 units = 1 dozen. 
12 dozen = 1 gross. 

CIRCULAR MEASURE. 

60 seconds (") — 1 minute. 

60 minutes (') = 1 degree. 

360 degrees (°) = 1 circumference. 

90 degrees = 1 quadrant. 

TIME MEASURE. 

60 seconds = 1 minute. 

60 minutes = 1 hour 

24 hours = 1 day. 

7 days = 1 week. 

365 days, 5 hours, 48 minutes, 48 seconds = 1 year. 

HORSE POWER. 

33,000 foot pounds = 1 horse-power (H. P.), that is equals 
the work required to raise 33,000 pounds 1 foot high in 1 
minute. 



70 



LUDLOW TEXTILE ARITHMETIC 



WEIGHTS AND MEASURES. 
WIRE GAUGES. 



Gauges. 


Birmingham. 


American or Brown 
and Sharp 




inch. 


inch. 


0000000 






000000 






00000 






0000 


0.454 


0.46 


000 


0.425 


. 40964 


00 


0.38 


. 3648 





0.34 


. 32486 


1 


0.3 


0.2893 


2 


0.284 


0.25763 


3 


0.259 


. 22942 


4 


0.238 


0.20431 


5 


0.22 


0.18194 


6 


0.203 


0.16202 


7 


0.18 


0.14428 


8 


0.165 


0.12849 


9 


0.148 


0.11443 


10 


0.134 


0.10189 


11 


0.12 


0.09074 


12 


0.109 


. 08081 


13 


0.095 


0.07196 


14 


. 083 


0.06408 


15 


0.072 


. 05707 


16 


0.065 


0.05082 


17 


0.058 


0.04526 


36 


0.004 


0.005 



MISCELLANEOUS EXAMPLES. 
LESSON LXXVI.— Written. 

1. — A bale of jute measures, approximately, 50" high, 
20" wide, 20" thick. How many of these could be packed 
in a storehouse 100 feet long, 50 feet wide and 20 feet high, 
allowing a passageway of 2 ft. all the way around the inside 
of the building? 

2. — How many systems of preparing machinery could 
be placed in a room 198' X88' allowing 1450 sq. ft. of floor 
space for each system? 



LUDLOW TEXTILE ARITHMETIC 71 

3. — How many feet of telephone wire would be required 
for a single line between Ludlow and Worcester, a distance 
of about 48 miles? 

4. — How many storehouses could be erected on a lot of 
20 acres, allowing 100 square feet for each house? 

5. — How many gallons would a cistern hold that is 31 
feet high and 8 ft. square? 

6. — How many feet between the Ludlow and Spring- 
field Post Offices, a distance of 7.8 miles? 

7 rf — Which box is larger and how much, one 36" long, 28" 
deep and 27" wide the other 30" long, 30" deep and 26" 
inches wide? 

8. — If a man empties 24 boxes of rove, each box contain- 
ing 56 roves, and each rove weighing 2 lbs., in 10 hours, how 
many pounds would he have to lift in 55 hours? 

9. — How much floor space will be required for 30 spin- 
ning frames 27 ft., 6 in. long and 7 ft. wide allowing for 
alleyways 7 ft. wide? 

10. — How many gallons of water will be contained in a 
catch-basin 2 ft., 9 in. high, 65 in. long and 23 in. wide? 

11. — How many pounds pressure would be put on a re- 
taining roller by 39 weights each weighing 3 lbs., 8 ozs.? 

12. — How many bobbins would be required to fill a box 
2 feet, 6 inches long, 2 feet wide, and 2 feet, 4 inches deep, 
each bobbin measuring 6 inches in length and 3 inches in 
diameter ? 

13. — If one boiler consumes 18 c. w. t. of coal in 24 hours, 

how much coal would 3 boilers consume in a week of 6 days? 

14. — If a doffer lifts 2 lbs., 6 ozs. of yarn every time she 

doffs a spinning frame how much will she have lifted at the 

end of a week of 55 hours, doffing every 30 minutes? 

15. — At a dollar a cubic foot, what would it cost to con- 
struct a building 500 feet long, 72 feet wide, 80 feet high. 



ALLIGATION. 



Alligation is useful in mill work for finding the average 
price per pound of any of the products which are made from 
fibres of different qualities having different cost prices. 

Alligation treats of mixing different substances at differ- 
ent prices producing a compound of some intermediate 
quality or price. 



72 LUDLOW TEXTILE ARITHMETIC 

Example. What is the average cost per pound of a mix 
made up as follows: 

Five 400 lbs. bales of jute @ 4$/ per pound? 
Four 400 lbs. bales of jute @ 50 per pound? 
Two 400 lbs. bales of jute @ 11$/ per pound? 
Three 400 lbs. bales of jute @ 12$/ per pound? 

Solution: 

5X400 = 2000 2000 X 4= 8000 

4X400 = 1600 1600 X 5= 8000 

2X400= 800 800X11= 8800 

3X400 = 1200 1200X12 = 14400 



5600 lbs. 39200$/ 

39200 

- — = 70 average price per pound. 

5600 

Rule. To find the average cost per pound of a mix 
when the proportion of the materials mixed and their prices 
are given, divide the total value of the material mixed by 
the sum of the amounts put in and the quotient will be the 
average price per pound. 

LESSON LXXVIL— Written. 

1. — What is the price per pound of the following mix: 
2/10 jute at 80 per pound, 1/10 hemp at 14$/ per pound, 4/10 
tow at 3 1/2$/ per pound, and 3/10 flax at 90 per pound? 

2. — Find the cost per pound of a mix made up as follows: 
1/3 at 10$/ per lb., 1/9 at 14$/ per lb., and 5/9 at lip' per lb. ? 

3. — What is the cost per pound of the following mixture: 
1/8 tow at 70 per lb., 1/4 flax at 11$/ per lb., 1/2 hemp at 12$/ 
per lb., 1/8 flax at 16$/ per lb.? 

4. — What would be the average price per pound of the 
following mixture: 1/4 tow @ 80 per lb., 1/2 hemp @ 12$/ 
per lb., 1/4 flax @ 16/ per lb.? 

5. — Find cost per pound of a mix composed as follows: 
1/3 flax @ 12$/, 1/3 hemp @ 10$/, 1/6 tow @ 9$/, and 1/6 jute 
@ ll$z. 

Alligation Alternate is the process of mixing quanti- 
ties at different prices so as to obtain a mixture of a required 
intermediate price. 



LUDLOW TEXTILE ARITHMETIC 73 

Example. It is desired to mix jute costing 60 per pound 
with hemp costing 90 per pound so as to make a mix worth 
70 per pound. What proportions of the different materials 
will have to be used? 

. 06 2 jute 

0.07 

. 09 1 hemp 

Write the prices of the jute and hemp in a vertical column 
as above with the price' of the desired mix at the left. 

Wrjte the difference between the desired price of the mix 
and the price of the jute against the hemp, in this case 1, 
and the difference between the price of the mixture and that 
of the hemp against the jute, in this case 2, and these figures 
will represent the proportions in which each of those quanti- 
ties enter into the mixture. 

LESSON LXXVIIL— Written. 

1. — Mixing flax costing 15 cents a pound with hemp 
costing 10 cents a pound t what proportions of the materials 
will have to be used to form a mixture worth 12 cents a 
pound?" 

2. — With jute at 5 cents a pound and hemp at 15 cents a 
pound, what proportions will go together to make a mixture . 
worth 10 cents a pound? 

3. — What proportion of hemp at 8 cents a pound and flax 
at 14 cents a pound will go together to form a mixture worth 
12 cents a pound? 

4. — How will four kinds of fibre worth respectively 60, 
80, 130, 160, a pound, be mixed to make a mixture worth 
10^ a pound? 

5. — With hemp at 10^, flax at 12^, jute at 70, and tow 
at 50, what proportions will go together to form a mixture 
worth 90 a pound? 



PERCENTAGE. 



Percentage is used in mill work for a variety of purposes. 
It is used to find the loss by waste in the different processes 
of manufacture. It is used again to find the relative cost of 
manufacture in the different departments through which 
the material passes, and it may be used to calculate the 
different speeds at which different parts of the machinery 
are running. 



74 LUDLOW TEXTILE ARITHMETIC 

Per Cent, means per hundred. 

The sign % means per cent. 

The Per Cent, of a number is the result obtained by- 
taking a stated number of hundredths of it. 
1% of a number is 1/100 of it. 
2% of a number is 2/100 of it. 
9% of a number is 9/100 of it 

The Percentage is the amount computed on the given 
number thus the percentage on $500 at 5% is $25. 

The Base of percentage is the number on which the per- 
centage is computed. 

Thus $500 is the base on which the percentage is com- 
puted in the foregoing example. 

When the per cent, is expressed by a decimal of more 
than two figures the figures after the second place must be 
taken as parts of 1%. Thus 0.125 means 12 5/10% or 12 
1/2%. 

The Rate Per Cent, being a certain number of hun- 
dredths may be expressed decimally or by a common fraction 
as is shown in the following table : 

1% = 0.01 = 1/100 

2% = 0.02 = 2/100 = 1/50 

10% =0.10= 10/100 = 1/10 

50% = . 50 = 50/100 = 1/2 

100%, = 1.00 = 100/100 = 1 

125%, = 1.25 = 125/100 = 1 1/4 

6|% =0.0625 = 61/100 = 1/16 

124% =0.125 = 12J/100 = 1/8 



TO EXPRESS A COMMON FRACTION AS A RATE 
PER CENT. 

Example. Express 4/5 as a rate per cent. 
1 = 100/100 = 100% 
4/5 = 4/5 X 100/100 = 80/100 
= 80% 

Rule. Divide 100 by the denominator of the fraction 
and multiply the quotient by the numerator. 



IvTJDLOW TEXTILE ARITHMETIC 75 

LESSON LXXIX.— Written. 

Express the following in rates per cent.: 

1.— 1/4 

2.— 1/3 

3.— 1/2 

4.— 1/6 

5.— 1/8 

6.— 1/10 

7.— 5/16 

8*— 5/12 

9.-3/7 
10.— 4/9 
11.— 9/40 
12.— 23/16 

TO EXPRESS A DECIMAL AS A RATE PER CENT. 

Example. Express 0.3 as a rate per cent. 
0.3 = 0.30 = 30% 

Example. Express 0.5625 as a rate per cent. 
0.5625 = 0'. 56 25/100 = 56)<%. 

Rule. Write the decimal as hundredths and the num- 
ber expressing the hundredths is the rate per cent. If the 
decimal has more than two places the figures that follow 
the hundredths place signify a part of 1%. 

LESSON LXXX.— Written. 

Express as a rate per cent.: 

1.— 0.46 

2.— 0.38 

3.— 0.07 

4.— 0.9 

5.— 0.26 

6.-7.79 

7.-6.48 

8.— 0.0005 

9.— 10.04 
10.— 0.00625 
11.— 0.125 
12.— 0.0375 



76 LUDLOW TEXTILE ARITHMETIC 

TO FIND THE PERCENTAGE HAVING THE BASE AND 
THE RATE GIVEN. 

Example. — Base = 250 and rate per cent. =4%, find the 
percentage. 

4% of 250 = 250X0.04=10.00 = 10 

or 
4% of 250 = 250 
.04 



10.00 
Rule. Multiply the base by the rate and point off two 
places to the left in the result. 

LESSON LXXXL— Written. 

1.— What is 6% of 250 pounds of jute? 

2.— Find 8% of 400 lbs. of jute. 

3.— Find 12i^% of 500 lbs. of flax. 

4. — 16K% of 1200 lbs. of hemp is equal to what? 

5.— What is 25% of 5000 lbs. of jute? 

6.— Find 33^% of 6600 lbs. of hemp. 

7.— Find 40% of 7000 lbs. of jute. 

LESSON LXXXIL— Written. 

1. — In a spinning room having 3600 spindles, making 
both hemp and jute yarns, 40% are spinning jute yarn. 
Find the number of spindles running on hemp. 

2. — The waste from a certain jute finisher card is averag- 
ing 4%. How many pounds of waste would this card make 
in a day's work of 3250 pounds? 

3. — From a certain grade of American hemp the loss by 
waste in the process of manufacture is as follows: 

Softening, 3% 

Breaker card, 16% 

Finisher card, 7% 

Drawing and roving, 3% 

Spinning and reeling, 3% 

What is the total loss in pounds per ton? 

4. — In the manufacture of twines from Italian hemp the 
waste is computed to be about 20%. On this basis how 
many pounds of finished twines can be made from, a lot con- 
taining 37„4> tons? 

5. — If 80 bagging looms are turning off 3860 yards of 
cloth in a week, and it is intended to increase the production 
by 20%, how many more looms will have to be installed to 



LUDLOW TEXTILE ARITHMETIC 77 

get this production, and what will be the total number of 
yards of cloth produced weekly when these additional looms 
are running ? 

TO FIND THE RATE PER CENT., THE BASE AND PER 
CENTAGE BEING GIVEN. 

Example. $1 is what per cent, of $4? 

1 X 100 = 100 

100-^-4 = 25% 
Example. 8 is what % of 200? 

8X100 = 800 

800^-200 = 4% 
Rule. Multiply the Percentage by 100 and divide the 
product by the Base. 

LESSON LXXXIII.— Written. 

1.— What per cent, of $350 is $19? 

2.— What per cent, of $340 is $14? 

3. — What per cent, of 250 pounds is 5 pounds? 

4 — What per cent, of 270 pounds is 16 pounds? 

5. — From a sample lot of American hemp containing 1,200 
pounds, there is a loss by waste and so forth of 444 pounds. 
Find the per cent, of loss in manufacture. 

6. — From a sample test of 1,000 pounds of Italian hemp, 
770 pounds of yarn are produced. What is the per cent, 
of loss? 

7. — The number of children attending the schools of 
Ludlow from the year 1896 to 1908 is as follows: 

Year. No. of children. Per cent, increase or decrease. 



1896 


315 


1897 


324 


1898 


325 


1899 


362 


1900 


448 


1901 


487 


1902 


523 


1903 


515 


1904 


470 


1905 


578 


1906 


607 


1907 


605 


1908 


723 



Show the annual rate of increase or decrease. 



78 LUDLOW TEXTILE ARITHMETIC 

TO FIND THE BASE WHEN THE PERCENTAGE 
AND RATE ARE GIVEN. 

Example. $6 is 3% of what? 
$6 X 100 = $600 

-r-3 



Rule. Multiply the percentage by 100 and divide the 
product by the rate. 

LESSON LXXXIV.— Oral. 

1.— $9 is 4% of what? 
2.— $37.50 is 3% of what? 
3.— 12 is 3% of what? 
4.— 37 y 2 is 6% of what? 
5.— 33 is 3/8% of what? 
6.-75 is 25% of what? 
7.— 120 is 20% of what? 

LESSON LXXXV.— Written. 

1. — From a sample of Italian hemp there are 330 pounds 
of waste, which are 23% of the whole. How many pounds 
were in the sample? 

2. — From a sett of jute laps there are 10 pounds of flyings 
which are 4% of the whole. What is the weight of the sett 
of lapps ? 

3. — A lot of American hemp loses 40% in being manu- 
factured into twine. The total amount of loss is 4,800 
pounds. How many pounds did the lot contain? 

4. — A boy is absent from work, in one week, 20% of the 
running time, and loses $1.20 in wages on that account. 
What are his wages per week? 

5. — In 1904 the number of children attending school in 
Ludlow was 470. Since that time the number has increased 
about 50%. How many are attending the schools now? 

6. — If a system of preparing machinery is turning off 
19,350 pounds weekly, and it is decided to increase the pro- 
duction by 5%, how many pounds per week will be produced ? 

LESSON LXXXVL— Written. 

1. — From a lot of jute there was made 1320 pounds of 
waste which was at the rate of 5.5%. How many 400 lb. 
bales did the lot contain? 



LUDLOW TEXTILE ARITHMETIC 79 

2. — -On May 1, 1900, the assessed valuation of personal 
estate in the Town of Ludlow was $458,138. On May 1, 
1907, the valuation was $1,602,769. Find the per cent, of 
increase during the seven years. 

3. — According to the assessors' report for the Town of 
Ludlow in the year 1900, the valuation of the real estate 
subject to taxation was $1,175,021. From the same source it 
is found that in 1907, the valuation of real estate had in- 
creased to $1,869,705. Find the per cent, of increase. 

4.— From the foregoing two examples find the per cent, 
of increase in the total valuation of both personal and real 
estate in Ludlow from 1900 to 1907. 

5. — According to the state census of 1905 the population 
of Ludlow was 3,881. In 1895 the census gives the popu- 
lation of the town as 2,562. Find the per cent, of increase 
during the ten years. 

SUMMARY OF RULES. 

Percentage =baseXrate. 
Base = percentage -r- rate. 
Rate = percentage -4- base. 
Amount = base X (1 + rate) . 
Difference = base X (1-rate). 
Base = amount -f- (1 -1-rate). 
Base = difference -4- (1-rate). 



SQUARE ROOT. 

Square Root is used in mill work to find the proper 
degree of twist to be put in the different sizes of yarns and 
ropes. 

The square root of a number is one of the equal factors 
of the number. The sign \/ , called the radical sign 

indicates square root, thus \/81 means the square root of 81 
or 9. 

(Square root may also be expressed by the fraction ^ 
written to the right and slightly above the number, 81). 

Examples of square roots. 

Numbers 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 

Square roots of 

the above 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 



80 LUDLOW TEXTILE ARITHMETIC 

In these cases all of the square roots are whole numbers. 
It is obvious that the square root of any number between 1 and 
4, or between 4 and 9, and so on, will be a mixed number 
whose value, in the first case will be between 1 and 2, in the 
second case between 2 and 3, and so forth. Thus a square 
root is not necessarily always a whole number. 

27X27 = 729 or it may be stated 
27 is the square root of 729 or, inversely 
V729 = 27 
Now 27 = 20 + 7 
then 27X27= (20 + 7) X (20 + 7) = 

20X(20 + 7)+7x(20 + 7) = 
20X20 + 20X7 + 20X7 + 7X7 = 
20.X 20 + 2X20X7 + 7X7 

Thus a number may be separated into two figures, one, 
the value of the tens, and the other, the value of the units 
and the product of the original number by itself (the square 
of the number) is equal to the square of the number repre- 
senting the tens plus twice the product of the tens and the 
units plus the square of the units. 

The extraction of the square root of a number, which 
is the reverse of this process, is based upon the above. 

It has been shown that the square root of any number 
from 1 to 99, inclusive, lies between 1 and 10. It can also 
be shown that the square root of any numbei from 100 to 
9,999 lies between 10 and 100, and so on. Thus the square 
root of any number expressed by one or two figures will have 
but one figure; the square root of any number of three or 
four figures will have two figures, and so on. Stated in an- 
other way, the square root of any number will have one 
figure for each two of the original number and one more 
in case of there being an odd number of figures in the original 
number. 

For example, the square root of 3,349 will have two 
figures; the square root of 7,946,347 will have four figures. 

This rule also applies to decimals, the square root contain- 
ing half as many decimal places as the original figure, with 
one extra if there is an odd number of decimal places in the 
original number. 

For example, the square root of 4,729.4345 will have 
two figures in the whole number and two additional decimal 
places. 



LUDLOW TEXTILE ARITHMETIC 81 

Advantage may be taken of the above when the square 
root of a number that is not a perfect square is being found 
for as many ciphers in the decimal places may be annexed as 
is desired in finding an approximate square root. 

In finding the square root of a fraction, the fraction may 
be reduced to a decimal or the square root of both numerator 
and denominator may be extracted. 

RULE FOR EXTRACTING THE SQUARE ROOT. 

Mark off the number into groups of two figures, com- 
mencing at the decimal point. 

Find the greatest square in the first group at the left 
and subtract it from the group. Place the square root of 
this square at the right as the first figure in the square root 
of the original number. Bring down the next group or 
pair and annex this to the remainder for a new dividend. 

Double the root obtained, annex a cipher, and use as a 
trial divisor. . The quotient, or a smaller number, will be the 
next figure in the root. 

Add this figure to the trial divisor for a complete divisor 
and multiply by the last figure. 

Subtract the product from the dividend; bring down the 
next pair of figures and annex them to the remainder. 

Continue in this way until all of the pairs of figures have 
been brought down. The result will be the square root. 

Example. Find the square root of 21,224,449. 

Marking off in pairs 

commencing at the right 

we have four. Four fig- 

21 22 44 49 ( 4607 sq. rt. ures will be the root. 

16 . The highest square in the 

86 ) 5 22 first pair of figures at the 

5 16 right is 16. Subtracting 

this from 21 the remain- 

9207 ) 6 44 49 der is 5. The square root 

6 44 49 of 16 is 4. This is the 

first figure in the answer. 

Bringing down the next 
group, 22, and annexing 
to the remainder, 5, the 
new dividend is 522. 



82 LUDLOW TEXTILE ARITHMETIC 

Doubling the root 4 and annexing a cipher the trial 
divisor is 80. 80 will go in 522 6 times. Add 6 to the 80 
making 86 for a complete divisor. Multiply by 6 and sub- 
tract this product from the dividend. The remainder will 
be 6. 

Bring down the next pair, 44. Double the root obtained, 
46, and annex a cipher for a new trial divisor. This will give 
920. As it is larger than the dividend 644, bring down the 
next pair, 49, and annex to the former dividend making the 
new dividend, 64,449. Place a cipher in the root as it was 
found that the trial divisor went into the dividend no times. 

Double the root 460, for a new trial divisor and annex a 
cipher. The trial divisor will be 9,200 and the dividend 
64,449. The divisor will be contained about 7 times Annex 
the 7 to the root and add 7 to the "trial divisor. The com- 
plete divisor 9,207 will now he contained exactly 7 times. As 
there is no remainder, the square root of 21,224,449 is 4,607. 

This may be proved by squaring 4,607, the product 
equaling the original number if the work is correct. 

LESSSON LXXXVII.— Written. 

Find the square root of: 

1.-123,201 8.-729 

2.-8,046.09 

3.-49,112,064 961 

4.— 110.25 9.-225 

5.-96,275,344 — 

6.-43.267 4489 

7.— 0.0346 10.— 532 

769 

Note — For practical examples see chapter on General 
Mill Work. 



GENERAL MILL WORK. 



TO FIND THE CIRCUMFERENCE, HAVING THE 
DIAMETER. 

Example. What is the circumference of a roller whose 
diameter is 4 inches? 

4"X 3. 1416 = 12.5664 

= 12.6" approximately 
Rule. Multiply the diameter by 3.1416. 



LUDLOW TEXTILE ARITHMETIC 83 

TO FIND THE DIAMETER HAVING THE 
CIRCUMFERENCE 

Example. Find the diameter of a roller whose circum- 
ference is 12.5664 inches. 

12.5664" -3. 1416 = 4" 

Rule 1. Divide the circumference by 3.1416. 

Another way to find the circumference is by proportion. 
This is an approximate method but the results are fairly 
accurate. The ratio of the diameter of any circle to its 
circumference is about 7 : 22. 

Therefore the proportion is 7 : 22 :: 4 : ? 

LESSON LXXXVIIL— Written. 

1 — Find the circumference of a roller whose diameter is 
2% inches. 

2. — What is the diameter of a roller whose circumference 
is 22 inches? 

3. — What is the circumference of a bicycle wheel whose 
diameter is 28 inches? 

4. — Find the circumference of a 5 foot cylinder. 

5. — Find the diameter, in inches, of a reel 23^ yards in 
circumference. 

6. — What is the diameter (working) of a fluted roller whose 
outer circumference is 8 inches? 

Note — -The circumference of fluted rollers such as are 
used in the wet spinning department is found by multiplying 
the diameter by 3.4. 

TO FIND LENGTH OF BELL ON LAPPER, CARD, 
REEL, OR OTHER MACHINES. 

Example. What is the length of bell on a lapper whose 
callendar roller measures 22^ inches in diameter and 
whose bell wheel has 84 teeth? 
22><X3.1416 = 70 
• 70X84 = 5880 

5880 -36 = 163 K yards 
Rule. Multiply the number teeth in bell wheel by the 
circumference in inches of the roller and divide by 36. (Num- 
ber of inches in 1 yard.) 

Note. — This is with a single worm. With a double 
worm the result would have to be divided by 2 to obtain 
the correct number of yards. 



84 LUDLOW TEXTILE ARITHMETIC 

LESSON LXXXIX.— Written. 

1. — What would be the length of bell on the lapper in the 
foregoing example with a 90 teeth bell wheel ? 

2.— With a 72 bell wheel? 

3.— With an 80 bell wheel? 

4. — With an 84 bell wheel and a drum 24 inches in di- 
ameter? 

TO TRACE THE SPEED OF ANY MACHINE FROM 
MOTOR OR DRIVING SHAFT. 

Example 1. Find the R. P. M. of spindles in No. 16^ 
Dept. when the speed of the cylinder is 411 R. P. M., the 
diameter of the cylinder is 10 inches, and the diameter of 
the wharve is 1% inches. 

411X10^1.5 = 2,740 R. P. M. 

Rule. Multiply the R. P. M. of cylinder by its diameter 
and divide the product by the diameter of the wharve. 

Example 2. What are the R. P. M. of a cylinder with 
shaft running 320 R. P. M., an 18" drum on shaft, and a 
14" pulley on the cylinders? 

320X18-14 = 4113/7 R. P. M. 

Rule. Multiply the R. P. M. of the driving shaft by the 
diameter of the drum and divide this product by the diameter 
of the pulley on cylinder end. 

Example 3. — Find the R. P. M. of a card cylinder in 
No. 72 Dept. when the speed of the motor is 800 R. P. M., 
the diameter of the motor drum 18 inches, the diameter of 
the motor pulley 52 inches, the diameter of the card drum 
18 inches, and the diameter of the card pulley 24 inches. 

By cancellation 

100 

2W 9 3 

px^xw 

=208 R. P. M., approx. 

13 

General Rule. Multiply the R. P. M. of the motor 

or driving shaft by the product of the diameters of the 

drivers and divide this product by the product of the diameters 

of all the followers. The result will be the required R. P. M. 



LUDLOW TEXTILE ARITHMETIC 85 

Note: — To distinguish between drivers and followers, 
call the drivers Drums, and the followers, Pulleys, as 
is the usual practice in the mills. 

TO TRACE SPEED OF SHAFT FROM DRAWING 
ROLLER OF SPINNING FRAME. 

Example. What is the speed of a driving shaft when 
the drawing roller is running 40 R. P. M., roller wheel 120 
teeth, twist pinion 36 teeth, stud wheel 90 teeth, cylinder 
pinion 29 teeth, and the pulley 14" diameter and the drum 
18" diameter. 

40X120X90X14 

=322 R. P. M. 

36X29X18 

Rule. Multiply the R. P. M. of drawing roller by the 
number of teeth in its wheel and multiply this product by 
the number of teeth in the crown wheel, and also by the 
diameter of the pulley for a dividend. For a divisor multiply 
the product of number of teeth in twist and cylinder pinions 
by diameter of drum on shaft. Divide the latter by the 
former and result will be R. P. M. of driving shaft. 

Note : — It is usually more convenient to trace the speed 
from one point to another by cancellation. 

LESSON LXXXX.— Written. 

Show the different speeds on a lathe with the driving 
cone running 175 R. P. M. 

Diam. steps on driving cone Diam. steps on follower 

10 inches 4 inches 

8 inches 6 inches 

6 inches 8 inches 

4 inches 10 inches 

TO FIND THE SURFACE SPEED OF A CYLINDER 
ROLLER OR PULLEY. 

Example. Find the surface speed in feet per minute 
of a card cylinder 48 inches in diameter running 210 R. P. M. 

48" -s-12 = 4 

4X3.1416 = 12.56 

12.56X210 = 2,638 feet per minute. 
Rule. Multiply the R. P. M. by the circumference. 



86 LUDLOW TEXTILE ARITHMETIC 

LESSON XCI.— Written. 

1. — Find the surface speed of a 14" (diameter) stripper 
in an American breaker card running 130 R. P. M. 

2. — What is the surface speed of the 9" (diameter) workers 
in this machine, when they are running 13 R. P. M.? 

3. — The diameters of the doffers at the points of the 
pins in some English breaker cards are about 20 inches. The 
pins in these Ludlow breaker cards is about 20 inches. The 
R. P. M. are 20. At what velocity will the jute travel pass- 
ing over the pins of the doffers ? 

4. — Find the R. P. M. a roller 4 inches in diameter would 
have to make to deliver as much sliver as the foregoing 
doffers. 

5 — Find the velocity in feet per minute of a belt running 
over a drum 60 inches in diameter while the shaft is running 
320 R. P. M. 



DRAFTING. 



The Draft of a machine is simply the amount of "draw" 
put on the fibre between the receiving and delivering rollers, 
To find the draft, it is possible to obtain the speeds of these 
two rollers and multiply these by their respective diameters 
times 3.1416 (circumferences) and thus find how much one 
receives and the other delivers in a minute. 

The usual method foi> finding the speed of any roller 
is to multiply the R. P. M. of the shaft by diameter of the 
driver and divide this result by the diameter of the follower, in 
order to trace the speed to any roller. In the rules given 
here, the speed of the shaft is not taken into consideration 
in finding the speed of either roller, consequently the same 
difference is kept proportionally between the two rollers. In- 
stead of using the circumferences, the diameters of both 
rollers are used, thus keeping the proportion the same. In- 
stead of multiplying and dividing as is usually the case in 
tracing the speed, the divisors and dividends are all collected 
together and worked out by cancellation as much as possible. 

As stated before draft is the ratio between the rates of 
feed and delivery. 



LUDLOW TEXTILE ARITHMETIC 87 

TO FIND THE DRAFT ON A ROVING, DRAWING, 
OR SPINNING FRAME. 

Example. Find the draft on a roving frame geared as 
follows : 

Boss roller pinion 38 

Draft wheel 36 

Back shaft pinion 24 

Stud wheel 70 

Stud pinion 24 

Back roller wheel 70 

Back roller diameter 2" 

Boss roller diameter 2%" 

38X24X24X2 

=9 draft 

36X70X70X2^ 

Rule. Multiply the product of all the drivers by diame- 
ter of receiving roller for a divisor; multiply the product of 
all the followers by the diameter of delivering roller for a 
dividend. Divide the latter by the former and the quotient 
will be the draft of the frame. 

By dividing the draft produced, 9 in this case, into 36, 
the number of teeth in the draft wheel, the constant number 
for draft of this frame is obtained, that is 9 into 36 = 4. 

To produce any draft on this machine multiply draft 
required by 4. The result will be the pinion required. 

TO FIND THE DRAFT ON A BREAKER CARD. 

Multiply the number of teeth in the pinion on the end of 
the delivery roller by the pinions driving the feed roller and 
this product multiplied by the diameter of the feed roller 
for a divisor. 

For a dividend, multiply the driven wheel between the 
delivery and the feed by the diameter of the delivering 
roller. Divide the former into the latter and the result 
will be the draft of the breaker. 

Example. Find draft of gearing in actual operation in 
some English manufactured carding machines. 

64X 17X 24X10 

=12.8 draft 

58X120X120X 4 



88 LUDLOW TEXTILE ARITHMETIC 

Example. Find wheel for 10 of a draft. 
10X4 = 40 wheel required 

For a draft of 8. 

8X4 = 32 wheel. 

Note: — The change pinion for draft on a roving frame 
is usually on the back shaft. 

The draft on a second drawing frame is found in the same 
way as for a roving frame. 

Example. Drawing Frame. 



Boss roller pinion, 


53 


Back shaft wheel, 


34 


Back shaft pinion, 


25 


Back stud pinion, 


25 


Back stud wheel, 


68 


Back roller wheel, 


69 


Back roller diameter, 


2" 


Boss roller diameter, 


2%" 


53X25X25X2 






- = 6 draft. 



34X68X69X2^ 
Note: — The change pinion for the draft on our drawing 
frames is on the end of the drawing rollers. 

The draft of a push bar drawing is a little more difficult 
to work out, but it is calculated in exactly the same manner 
as the draft on any other machine. The difference between 
the number of inches taken in by the receiving roller and 
the number of inches delivered by the roller which delivers 
the fibre constitutes the draft. 

Example of the draft gearing on latest style of push bar 
drawings. Find draft. 

64X18X24X40X2 

= 5 

60X80X22X42X2^ 
On another style of push bar frame the gearing is : 
48X20X27X42X39X2 

= 5.1 

52X84X42X22X43X23^ 
On an English push bar frame the draft gearing now 
working is 

48X18X28X2 

= 5.1 

42X54X43X2^ 



LUDLOW TEXTILE ARITHMETIC 89 

TO FIND DRAFT OF A ROLLER CARD. 

The draft of a roller finisher card is found by multiplying 
the diameter of the feed roller by the product of all the 
driving wheels between the feed and the delivery for a divi- 
nend; for a divisor find the product of all the followers 
between the feed and the delivery and multiply by the diam- 
eter of the delivery roller for a divisor. 

By dividing the latter into the former the theoretical 
draft of the card is obtained. The actual draft of the card 
may »ot be quite as long as the figures show because the 
diameter of the feed roller is taken at the bottom of the 
pins while the fibres do not all go to the bottom of them. 

Example 1. Find draft of finisher card with draft 
gearing as follows: 



Feed roller wheel, 


96 


Draft pinion, 
Stud wheel, 


24 
96 


Stud pinion, 
Second stud wheel, 


32 

104 


Delivery roller pinion, 
Diameter feed roller, 


75 
4" 


Diameter delivery roller, 


4" 


96X96X104X4 


= 16.6 draft 


24X32X75X4 



This means that the sliver delivered is 16.6 times as 
long as the sliver that is received. 

Example 1. Find by proportion the draft pinion re- 
quired to have this machine on a draft of 20. 

Example 2. And the draft pinion for a draft of 12. 
On a Fairbairn Knife Card the particulars of the draft 
will run as follows: 

Feed roller wheel, 120 

Draft pinion, 44 

Inside stud wheel, 72 

Delivery roller pinion, 16 

Diameter feed roller, 3" 

Diameter delivering roller, 4" 

Then 120X72X4 

=16.36 draft. 

44X16X3 



90 LUDLOW TEXTILE ARITHMETIC 

Example. Find the draft with a 36 draft pinion on the 
card. 

2. . What draft pinion would be required to produce 
a draft of 18 on this card? 



SPINNING. 



TO FIND THE DRAFT ON A SPINNING FRAME. 

Rule. Multiply the number of teeth in stud wheel 
by number of teeth in top roller wheel and multiply this 
product by diameter of drawing roller for a dividend. Mul- 
tiply the diameter of top roller by the number of teeth in 
draft pinion and then multiply this product by the number 
of teeth on inside stud pinion for a divisor. The number of 
times this divisor is contained in the dividend is the draft 
of the frame. 

Example 1. Find the draft on a Z}4," spinning frame 
geared as follows: 

Draft pinion, 36 teeth 

Stud wheel, 80 " 

Inside stud pinion, 30 " 

Top roller wheel, 75 " 

Diameter top roller, 23^" 

Diameter drawing roller, 4" 

80X75X4 

= 8.8 draft 

36X30X2^ 

Example 2. With the draft gearing on a 4" as follows, 
find the draft. 

Draft pinion, ' 37 

Stud wheel, 80 

Inside stud pinion, 40 

Top roller wheel, • - 75 
Diameter top roller, 2^" 

Diameter drawing roller, 4" 

80X75X4 

— =6.4 draft 

37X40X2^ 



LUDLOW TEXTILE ARITHMETIC 91 

Example 3. Find the draft of a 4^" spinning frame 
with the following gears: 

Draft pinion, 38 teeth 

Stud wheel, 60 " 

Inside stud pinion, 40 " 

Top roller wheel, 80 " 

Diameter top roller, 2%" 

Diameter drawing roller, 4" 

60X80X4 

= 5.6 draft 



38X40X23^ 




Example 4. On a 5" spinning 


frame the draft gearing 


.ild run at follows: 




Draft pinion, 


56 teeth 


Stud wheel, 


50 " 


Inside pinion, 


45 " 


Top roller wheel, 


80 " 


Diameter top roller, 


2H» 


Diameter drawing roller, 


5" 


Find the draft. 




50X80X5 




= = 3.17 


draft 



56X45X2^ 

TO FIND THE CONSTANT NUMBER FOR DRAFT IN 
A SPINNING FRAME. 

Rule. Multiply number of teeth in stud wheel by 
number of teeth in top roller wheel and by diameter of 
drawing roller for a dividend; then multiply number of 
teeth in stud pinion (inside) by diameter of receiving roller 
for a divisor, or, use the figures for finding the draft with 
the draft pinion omitted. 

Example 1. To find the constant number of the 33^" 
spinning frame geared as shown in the example for finding 
the draft. 

80X75X4 

= 320 = constant number 

30X2^ 
Example 2. Find the constant number for draft on 
a 4" spinning frame. 

80X75X4 

= 240 constant number 

40X2.5 



92 LUDLOW TEXTILE ARITHMETIC 

Example 3. Find the constant number for draft on a 
4/^" spinning frame. 

60X80X4 

= 216 constant number 

40X2.5 
Example 4. On a 5" spinning frame find the constant 
number for draft. 

50X80X5 

=177 constant number 

45X2.5 

TO FIND THE PINION FOR ANY DRAFT HAVING 
THE CONSTANT NUMBER. 

Rule. Divide the draft required into the constant 
number ; the result will be the number of teeth on the pinion 
to be put on the spinning frame to produce the required 
draft. 

Example. What pinion would be required to produce 
a draft of 8 on a 3^2" frame, the constant number being 320? 
320 -f- 8 = 40 teeth on draft pinion 

TO FIND DRAFT A CERTAIN PINION IS PRODUC- 
ING BY HAVING CONSTANT NUMBER. 

Rule. Divide draft pinion into constant number. 

Example. What draft is on a certain frame whose 
constant number is 320 and which is running with a 40 
draft pinion? 

320-^-40 = 8 draft 

For additional examples in drafting, vary draft pinions 
on different machines. 



TWISTING. 



The twist in yarn is calculated by the turns per inch. 

(The turns per inch means the number of revolutions 
the Flyer makes while the rollers are delivering one inch 
of yarn.) 

To find the number of turns per inch which any machine 
is putting into the yarn, divide the R. P. M. of the spindles 
by the number of inches that the rollers deliver per minute; 
the result will be the turns per inch. 



LUDLOW TEXTILE ARITHMETIC 93 

For example, the twist going in the yarn on a spinning 
frame, with the rollers delivering 564 inches per minute 
and the spindles running at the rate of 1920 R. P. M., would 
be 1920^-564 = 3.4 turns per inch. 

The twist of a frame can be found without having the 
R. P. M. of any part of the machine by getting all the divisors 
and dividends together and working out the problem as 
far as possible by cancellation. 

Rule. Multiply the number of teeth in roller whee 
by the number of teeth in stud wheel and this product by 
the diameter of cylinder, for a dividend; multiply the cir- 
cumference of drawing roller by diameter of wharves by 
twist pinion and by cylinder pinion for a divisor. 

Example. Find the twist of a spinning frame geared 
as follows: 

Roller wheel, 96 

Stud wheel, 136 

Diameter of cylinder, 10 

Cylinder pinion, 24 

Twist pinion, 56 

Circumference of roller, 14. 1" 

Diameter wharves, 2" 



96X136X10 
24X56X14.1X2 



5.2 turns per inch 



Example. Find the turns per inch of yarn spun on a 4" 
frame with the twist geared as follows: 



Roller wheel, 


120 


Stud wheel, 


90 


Diameter cylinder, 
Cylinder pinion, 
Twist pinion, 
Roller circumference, 


10 
28 
35 
12.56 


Diameter wharve, 


1.75 


120X90X10 


— = 5 turns per inch 



28X35X12.56X1.75 



94 LUDLOW TEXTILE ARITHMETIC 



Example 4. What would 


be 


the twist of a spinning 


frame geared like this? 






Roller wheel, 




150 teeth 


Stud wheel, 




112 " 


Diameter of cylinder, 




10" 


Cylinder pinion, 




44 teeth 


Twist pinion, 




39 " 


Circumference of roller, 




15. 7" 


Diameter of wharve, 




2 . 25" 


150X112X10 








— =tt V / +nr-n c nar 1-n/^n 



44X39X15.7X2.25 

A constant number for the twist of a spinning frame is 
. found by leaving the twist pinion out of the foregoing calcu- 
lations. For example, the constant number for a 3^ frame 
with the same gearing and diameter of roller as is given 
in example 4, would be 

120X90X10 197 

= = turns per inch 

T.P.X29xl.5xl2.56 T.P. 
Therefore the constant number for this machine would 
be 197. 

Example 5. Find the constant number for twist of a 
4" frame. 

120X90X10 

= 175 constant number 

28X1.75X12.56 

Example 6. Find the constant number for a 5" frame. 
150X112X10 

= 105 constant number 

44X2.25X15.7 

The average degree of twist required for flax, hemp, 
and jute yarns is found by extracting the square root of 
the leas per pound and multiplying this product by 2 for 
the number of turns per inch. 

Thus the number of turns per inch required for 4's lea 
yarn equal 2X\/4 = 2X2 = 4 turns per inch. 

And again the number of turns per inch required for 
25's lea would be 2X\/25 = 2X5= 10 turns per inch. 

For yarns numbered on the pounds per spyndle basis, it 
is usually ' taken that 3 lb. yarn requires 8 turns per inch. 
The number of turns per inch required for any other size 



LUDLOW TEXTILE ARITHMETIC 95 

of yarn may be obtained by multiplying the turns per inch 
for 3 lb. yarn by the \/3 and dividing by the \/ lbs. per 
spyndle of the yarn to be twisted. 

Thus, the twist required for yarn weighing 8 lbs. per 
spyndle at the rate of 8 turns per inch for 3 lb. yarn would 
be 

SXv/3 64X3 

= — = a/24 = 4.9 turns per inch 

V8 8 

The reason that the square root of the size of the yarn 
is introduced into the twist calculations is that the twist 
should vary inversely as the diameter of the thread and 
that the diameter of the thread varies inversely as the square 
root of the leas per pound or directly as the pounds per 
spyndle. 

From this reasoning we get a constant number for finding 
the turns per inch by multiplying 8, the turns per inch, for 
3 lb. yarn, by 1.73, the square root of 3, which equals 13.8. 

Thus for example to find the turns per inch required for 
8 lb. yarn, divide the square root of 8 into the constant 
number 13.8 which equals 13.8-^2.8 = 4.9 turns per inch. 
\/ = radical sign 



EXAMPLES IN TWISTING. 



LESSON XCIL— Written. 



1. — What would be the average twist or turns per inch 
for yarn weighing 9 pounds per spyndle? 

2. — Find the turns per inch for 6 lea yarn. 

3. — What is the twist for 14 pound yarn? 

4. — Make a sheet showing the twist required for each lea 
of yarn from l's to 10 's inclusive. 

5. — Make a list showing the turns per inch required for 
each one pound difference in weight per spyndle from 1 
to 6 lbs. 

Note. These examples are worked out on the average 
degree of twist that is required for ordinary yarns. In yarn 
wanted for special purposes, the twist will vary from this 
it may be 1.75 times the square root of the lea or it may be 
2.25 to 2.5 times the square root of the leas per pound. 



96 LUDLOW TEXTILE ARITHMETIC 

STRANDS. 

TO FIND PROPER DEGREE OF TWIST PER STRAND 

OF ROPE. 

The number of turns per foot twist required in a strand 
is found by dividing the number of the yarn used by the 
number of threads per strand, extracting the square root 
of this result and multiplying by 3.75. Thus the number 
of turns per foot twist required for a strand containing 40 
threads of number 20's rope yarn would be 

V20/40 X 3 . 75 = 2 . 7 turns per inch 
Note : — This twist must be put in the rope in the opposite 
direction to that in which the yarn was spun. 

TO FIND NUMBER OF THREADS PER STRAND OF 
ANY SIZE ROPE. 

The number of threads per strand for a 3 strand of any 
diameter may be found by dividing the product of the square 
of the rope's diameter and the number of the yarn by 0.81 
or 0.9 2 , 0.9 being the diameter of a rope 3 inches in circum- 
ference. 

The number of threads per strand for a 4 strand rope 
of any diameter is found by squaring the diameter of the 
rope and multiplying by the number of the yarn which is 
used and dividing the product thus obtained by 1.2. 

Thus the number of threads of 20's rope yarn required 
to form a strand for a three strand rope 2 inches in diameter 
would be 2 2 X 20 ^ 0.81 = 99. 

For a four strand rope 2 inches in diameter it would be 
2 2 x20 

= 67 

1.2 



CUTS PER SPINDLE. 



The yarn production is calculated by the average number 
of 300 yard cuts which each spinning spindle produces in 
a stated time. 

Rule: — Where the yarn is reeled, the number of reels 
multiplied by the number of cuts contained in each reel will 
give the total number of cuts, and this product divided by 



LUDLOW TEXTILE ARITHMETIC 97 

the number of spinning spindles running while this yarn 
was being produced, will give the average cuts per spindle 
for that particular space of time. 

Example. What are the average cuts per spindle for 
10 hours where the number of reels turned off is 600, each 
reel containing 100 cuts and the number of spindles in oper- 
ation, during that time was 3300. 

600 X 100 

= 18. 1 cuts per spindle per 10 hours 

3300 

Where the yarn is numbered on the pounds per spyndle 
basis the rule to find the cuts per spyndle is: 

Rule : — Divide the production by the pounds per spyndle, 
for the number of spyndles; multiply this quotient by 48 
(the number of cuts in a spyndle) for the total number of 
cuts ; divide this product by the number of spinning spindles 
in operation during the time this yarn was being produced, 
and the result will be the average cuts per spindle. 

Example. The average cuts per spindle from 200 spin- 
ning spindles which produce 1875 pounds of 15 lb. yarn 
in a day of 10 hours would be 

1875/15=125 
125X48 = 6000 
6000^200 = 30 

30 cuts per spindle per 10 hours, or 

3 cuts per spindle per hour 

which means that each spinning spindle in operation on that 
particular size yarn is spinning, on an average, 900 yards 
of yarn each hour. 



TO FIND THE CUTS PER SPINDLE PER HOUR. 

Multiply the number of spindles running by the number 
of hours that they have been in operation, the result will 
be the spindle hours and this product divided into the total 
number of cuts will give the cuts per spindle per hour. 



98 



LUDLOW TEXTILE ARITHMETIC 



Example. Find 


the cuts 


per spindle 


per hour fc 


following work : 








Kind 


Pounds 


Spvndles 


Cuts 


15 lbs. 


13,500 


900 


43,200 




Spindles 


Hours 


Spindle 
Hours 


Monday, 


300 


10 


3,000 


Tuesday, 


240 


10 


2,400 


Wednesday, 


300 


9 


2,700 


Thursday, 


210 


10 


2,100 


Friday, 


300 


10 


3,000 


Saturday, 


240 


5 


1,200 


Total spindle hours 


? 


14,400 


Total number of cuts = 43,200 


Q O11 + T-.0-T 


o-r-*i-nrl1o n&r 



Spindle hours = 14,400 

TO FIND THE NUMBER OF YARDS IN ANY WEIGHT 

OF YARN, THE POUNDS PER SPYNDLE 

BEING GIVEN. 

Rule. Divide pounds per spyndle into weight, multiply 
this product by 48 for cuts, and the number of cuts multi- 
plied by 300 gives the number of yards. 

Example. How many yards are contained in 12,000 
lbs. of jute yarn which weighs 15 lbs. per spyndle. 
12000/15 = 800 
800X48 = 38,400 yards 



LEVERS. 

TO FIND THE PRESSURE ON DRAWING ROLLER. 

Rule. Multiply length of lever in inches by the weight 
suspended in pounds, and this product divided by the length 
of fulcrum will give the pounds pressure on the roller. 

Example. To find weight on a pressing roller with 



Length of lever, 


12" 


Weight suspended,. 


14 lbs. 


Fulcrum, 


IX" 


14X12 




-134U lbs. 





\v 



LUDLOW TEXTILE ARITHMETIC 99 

On drawings and rovings where compound levers are 
used the pressure is found by multiplying the length ofthe 
two levers by the weight and dividing this by the product 
of the two fulcrums. 



Example. Second drawing fra: 


me. 


Top lever, 


9" 


Bottom lever, 


9" 


Top fulcrum, 


2" 


Bottom fulcrum, 


3y 2 " 


Weight, 


15 lbs. 


9X9X15 




= 173 4/7 lbs. 


pressure 



2X3.5 
ROVING FRAME. COMPOUND LEVER. 



Top lever, 


4" 


Bottom lever, 


5" 


Top fulcrum, 


1" 


Bottom fulcrum, 


2" 


Weight, 


12 lbs, 


4X5X12 




= 


120 lbs. 



2X1 
Note: — The word fulcrum in the above examples is a 
mill term meaning the short arm of the lever. 

TO FIND POUNDS PER SPYNDLE OF ROVE FROM 
WEIGHT OF 100 YARDS. 

Example. What are the pounds per spyndle of rove 
which weighs 10 ounces per 100 yards? 

10X9 = 90 lbs. per spyndle 

Explanation. The weight of 100 yards in ounces mul- 
tiplied by 9 is equal to the weight of a spyndle (14,400 yds.) 
divided by 16, the number of ounces contained in one pound. 

Thus 100 yards = 10 ounces in weight 
and 1 spyndle (14,400 yds.) = 1440 oz. in weight 
and 1440^16 = 90 lbs. 

Rule. Multiply weight of 100 yards of rove in ounces 
by 9; the result will be the pounds per spyndle. 



100 



LUDLOW TEXTILE ARITHMETIC 



LESSON XCIIL— Written. 

Find the pounds per spyndle of 100 yards of rove weighing 

1. — 5 ounces. 

2. — 12 ounces. 

3. — 7 ounces. 

4. — 16 ounces. 

5. — 10 ounces. 

TO FIND DRAFT TO BE PUT ON SPINNING FRAME 

HAVING POUNDS PER SPYNDLE OF 

ROVE GIVEN. 

Example. Find draft required to make yarn weighing 
14 lbs. per spyndle from rove weighing 81 lbs. per spyndle. 
81^14 = 5.8 draft 

Rule. Divide the weight of spyndle of yarn required 
into weight per spyndle of rove. 

LESSON XCIV.— Written. 

Find draft required to spin 

1.— 6 lbs. yarn from 63 lbs. rove. 



2.— 5 
3.— 15 
4.— 14 
5.— 30 
6.— 18 
7.-24 



63 

81 
81 
81 
90 
108 



TO FIND SPINNING DRAFT FOR ANY SIZE YARN 
FROM WEIGHT OF ROVE. 

Example. What draft would be required to spin yarn 
weighing 10 pounds per spyndle, from rove weighing 9 ounces 
per 100 yards? 

50 yards rove = 43^ ounces 
4373^2 grains = 1 ounce 
therefore 437 . 5 X 4 . 5 = 1968 . 75 grains 

and 50 yards, 10 lb. yarn = 243 grains 

1968. 75 -=-243 = 8.1 draft on spinning frame 
Rule. Divide weight of 50 yards of yarn required into 
the weight of 50 yards of rove. The quotient will be the 
draft to be put on the spinning frame. 



LUDLOW TEXTILE ARITHMETIC 



101 



LESSON XCV.— Written. 

Find draft required to make yarn from rove weighing 
10 ounces per 100 yards. 

1. — 50 yards yarn to equal 243 grains. 
2.— 50 " " " " 316 " 
3.— 50 " " " " 340 " 
4.— 50 " " " " 365 " 

TO FIND DRAFT TO SPIN ANY LEAS PER POUND 
HAVING WEIGHT OF ROVE GIVEN. 

Example. What draft is necessary to spin 3 lea yarn 
from rove weighing 9 ounces per 100 yards? 

9X3-5-5.3 = 5.1 spinning draft 

100 yards of rove equals 9 ozs. 

1 lea or 300 yds. =27 ozs. 

3 leas or 900 yds. of rove = 81 ozs. and the yarn required is 

3 lea or 900 yds. to the pound which equals 10 ozs. 

Therefore 81^16 = 5.1 

Placing this in the form of a fraction 
300X9X3 

= 5.1 

100X16 

Rule. Multiply weight of 100 yards of rove in ounces 
by leas per pound of yarn required. This product divided 
by 5 . 3 will give draft to be put on spinning frame. 

Note: — While the weight of rove and leas per pound are 
variable the number of yards in a lea (300), the ounces in 
a pound (16), and the number of yards of rove weighed 
(100), are unchanged. Therefore from these figures the 
following is obtained: 

100 X 16 h- 300 = 5 . 3 a constant number 



LESSON XCVL— Written. 

Find draft for spinning frame to spin 

1. — 3 lea from rove weighing 9 ozs. per 100 yards. 



2.-4 " " 


9 " 


" 100 " 


3.-6 " " 


6 " 


" 100 " 


4.-8 " " 


5i " 


" 100 " 


5.-2 " " 


10 " 


" 100 " 


6.— 1 " " 


12 " 


" 100 " 



102 



LUDLOW TEXTILE ARITHMETIC 



To obtain 

10 X 

100 X 
5X 
25 X 
125 X 
33KX 
12^ X 
5% of 
25% of 
50% of 
20% of 
33^% of 
66^% of 
12K% of 
16K% of 
9X 
HX 

To divide by 
10 
100 
5 

25 
125 
33^ 



SHORT PROCESSES. 

Move decimal point one place to right. 
Move decimal point two places to right. 
Multiply by 10 and divide by 2. 
Multiply by 100 and divide by 4. 
Multiply by 1000 and divide by 8. 
Multiply by 100 and divide by 3. 
Multiply by 100 and divide by 8. 
Divide by 10 and divide quotient by 2. 
Divide by 4. 
Divide by 2. 
Divide by 5. 
Multiply by 1/3. ' 
Multiply by 2/3. 
Multiply by 1/8. 
Multiply by 1/6. 

Multiply by 10, subtract multiplicand. 
Multiply by 1Q and add multiplicand. 

Move decimal point one place to left. 
Move decimal point two places to left. 
Multiply by 2 and divide by 10. 
Multiply by 4 and divide by 100. 
Multiply by 8 and divide by 1000. 
Multiply by 3 and divide by 100. 



Answers 



ADDITION. 

Lesson 2. Page 3-4. 

1.— 1665 7.-1,539,829 

2.-792 8.-800,686 

3.— 1208 9.-1,832 fbs. 

4.— 4018 10.— 1100 lbs. 

5.-102,332 11.— 21,852 lbs. 

6.-1,396,684 12.-44,696 lbs. 

JUTE YARN PRODUCTION SHEETS. 

Lesson 3. Page 4-5. 
1.-20,053 lbs. 4.-72,246 lbs. 

2.-19,155 lbs. 5.-42,715 lbs. 

3.-26,060 lbs. 6.-32,990 lbs. 

PREPARING ROOM SHEETS. 

Lesson 4. Page 6-7. 
1.-29,510 lbs. 233 doffs 

2.-9,300 lbs. 65 doffs 

3.-30,500 lbs. 214 doffs 

WEEKLY PRODUCTION SHEETS. 

Lesson 5. Page 7-8. 
1.-627,395 lbs. 
2.-621,915 lbs. 
3.-619,870 lbs. 

SUBTRACTION. 

Lesson 6. Page 8-9. 

1.— 571 8.-457 lbs. lost 

2.— 102 9.— 192 pupils increase 

3.— 7912 10 —665 men 

4.-657 11.— 974 people 

5.— 4177 12.-29,714 lbs. 

6.-37,987 13.— 143 lbs. 

7.-38,774 14.-27,941 people gain 



104 LUDLOW TEXTILE ARITHMETIC 

MULTIPLICATION. 

Lesson 7. Page 10-11. 



1.- 


—2403 






7.— $5.50 


2.- 


-57,817 






8. — 96 sides 


3.- 


—452,100 




9.— 918 help 


4.- 


-248,777,588 




10.-14,400 yds. 


5.- 


-19,250 lbs. 




11.-20,300 lbs. 


6.- 


-95,200 


yds. 


WAGE 


12.-1,500,000 lbs, 
LIST. 






Li 


ESSON 8. 


Page 11. 


1.- 


-$2 . 75 






9.— $7.15 


2.- 


-$3.30 






10.— $7.70 


3.- 


-$3 . 85 






11.— $8.25 


4.- 


-$4.40 






12.— $8.80 


5.- 


-$4.95 






13.— $9.35 


6.- 


-$5.50 






14.— $9.90 


7.- 


-$6.05 






15.— $10.45 


8.- 


-$6.60 




PRICE 


16.— $11.00 
LIST. 






Lesson 9. 


Page 11. 


1.- 


-$0.20 






11.— $2.20 


2- 


-$0.40 






12.— $2.40 


3.- 


-$0.60 






13.— $2.60 


4.- 


-$0.80 






14.— $2 . 80 


5.- 


-$1.00 






15.— $3.00 


6- 


-SI . 20 






16.— $3.20 


7.- 


-$1.40 






17.— $3 . 40 


8.- 


-$1.60 






18.— $3.60 


9.- 


-$1 . 80 






19.— $3.80 


10.- 


-$2 . 00 






20.— $4.00 








DIVISION. 






Lesson 10. 


Page 11-12. 


1.- 


-156 






6.-567 83/100 


2.- 


-2,552 5/7 




7.— 120 


3.- 


-9357 






8.-2768 8/11 


4.- 


-98,311 


3/8 




9.-21,120 8/13 


5.- 


-73,894 


2/5 




10.-59,768 12/15 



LUDLOW TEXTILE ARITHMETIC 



105 



LONG DIVISION. 

Lesson 11. Page 13. 



1.— 4158 
2.-274 46/345 
3.— 212 13/62 
4.-24,389 
5.-364 140/237 

Lesson 12. 
1.— 300 yards 
2^—48 cuts 

3.-10,800 ins. 300 yds. 
4.— 150 bales 



6.-22 303/685 
7.— 10 181/874 
8.-27 139/925 
9.-53 1393/7005 
10.— 8 1432/2469 
Page 13-14. 

5. — 5 turns per inch 
6.-253 1/3 spyndles 
7. — 25 cuts 
8.— $894 2860/3881 



MISCELLANEOUS EXAMPLES. 



Lesson 13. Page 

1.-17,280 pins 

2.-21,600 pins 

3.-42,224 pins 

4.-81,104 pins 

5. — 339 1/11 amperes 

6.— 740 yds. 

7.— 600 yds. 

8.— 380 yds. 

9.-284 yds. 
10.— 200 yds. 
11.— 150 yds. 
12.— 430 yds. 
13.— 380 yds. 
14.— 208 yds. 
15.— 1400 lbs. 
16.-40,000 lbs. 
17. — 141 revs. 
18.-57,600 revs. 
19.— $9.15 
20.— 60 hours 
21.— 20 hours 

ANALYSIS, 



14-15-16-17. 
22.— 340 grains 
23.-5,346,593 bales 
24.-48 leas 
25.— 320 lbs 
26.— 1000 spyndles 
27. — 437 1/2 grains 
28.— 2800 pounds 
29.-24 feet 
30.— 375 bobbins 
31.-23,809 11/21 ozs. 
32.-15^ 
33.— $14.40 
34.-800,000 pins 
35.-12,000 lbs. 
36.-37,800 pins 
37.-56 1/4 lbs. 
38.— 178 H. P. 
39.-2,520,000 yds. 
40.— 3055 5/9 spyndles 
41.— 165 gals. 



Lesson 14. 
1.— 110 doffs 
2.-33,000 lbs. 
3.— $2.25 
4.— 250 girls 
5. — 200 ounces 
6.— 3712 fallers 



Page 18. 

7.-225 bales 
8.-1,923,077 lbs. 
9.— $1,000,000 

10.— $125,000 

11.— 132 ft. 

12. — 6 cents 



106 LUDLOW TEXTILE ARITHMETIC 





Lesson 


15. 


Page 19. 


1.- 

2.- 

3.- 
4.- 
5.- 
6.- 


—12 1/4 cents 
-$8.60 
—10 men 
-6800 lbs. 
— 10 winders 
— 5 machines 




7.— 20' sides 

8.— 50 bales 

9.— $360 

10. — 36 boys 

11. — 56 boys 

12.— $36 



PRIME FACTORS. 

Lesson 17. Page 20. 

1.— 2-2-3-5 11.— 2-2-2-2-2-3-5 

2.-2-2-3-3 12.— 2-2-2-2-2-2-2-5 

3.— 3-13 13.— 2-2-2-2-3-3-5 

4.-2-3-7 14.— 2-2-5-5-7 • 

5.— 17 15.— 2-2-2-2-2-5-5 

6.-3-5-5 16.— 2-3-181 

7.-2-3-5-7 17.— 2-5-5-31 

8.-2-2-2-2-5-5 18.— 2-3-3-5-5-7 

9.-2-3-3-3-3-5 19.— 2-2-3-5-5-13 

10.— 2-2-3-3-3-7 20.— 3-3-3-5-11 

GREATEST COMMON FACTOR. 

Lesson 18. Page 21. 

1.— 12 6.-7 

2.— 18 7.-25 

3.— 11 8.-7 

4.— 12 9.— 13 ' 

5.— 16 10.— 60 

LEAST COMMON MULTIPLE. 

Lesson 19. Page 22. 



1.— 6 


6.— 20 


2.— 30 


7.— 30 


3.-56 


8.-24 


4.-36 


9.— 20 


5.— 12 


10.— 36. 


Lesson 21. Page 23. 


1.— 180 


.6.-96 


2.— 198 


7.-36 


3.— 140 


8.-45 


4.— 90 


9.— 6000 


5.— 144 


10.— 80 



LUDLOW TEXTILE ARITHMETIC 



107 



-2 

-22 

-3 



CANCELLATION. 

Lesson 22. Page 24. 

4.— 21 

5.— 1 



UNITED STATES MONEY. 

Lesson 23. Page 24-25. 



1.- 

2.- 
3.- 
4.- 
5.- 


-$10,000.40 

-$784,164 

-$5674.192 

-$3,000,000,003 

-$7,000,033 


6.— $807.10 
7.— $5023 . 768 
8.— $1,002 
9.— $27 . 305 




Lesson 24. 


Page 25. 


1.- 
2.- 

3.- 


-2246^ 
-4700^f 
-67 ; 590 mills 


4.-32,476 mills 
5.-66,432 mills 




Lesson 25. 


Page 25-26. 


1.- 
2.- 
3.- 


-$49.87 
-$7,658 
-$460.36 


4.— $9000.145 
5.— $10,049,015 




Lesson 26. 


Page 26. 


1.- 
2.- 
3.- 
4.- 


-$4.75 
-$5 . 00 
-$5 . 25 

-$5.75 


5.— $6.00 
6.— $6.25 
7.— $7.00 




DECIMALS. 




Lesson 28. 


Page 28-29. 


1.- 

2.- 
3.- 
4.- 
5.- 
6.- 
7, 
8.- 


-0.1 

-0.27 

-0.437 

-608.2701 

-941.32721 

-3.41159 

-19.083 

-0.001 


9.— 0.01763 
10.— 0.03003 
11.— 60.006 
12.— 170.704 
13.— 5.0856 
14.— 89.009 
15.— . 0378 
16.— 5.9764 



108 LUDLOW TEXTILE ARITHMETIC 

ADDITION OF DECIMALS. 

Lesson 29. Page 29-30. 

1.— 169.03 8.— 149.14438 

2.— 280.466 9.— 174.379779 

3.-224.9586 10.— 70.3932725 

4.— 189.28745 11.— 184.5376349 

5.— 512.362229 12.— 119.349505 

6.— 59.656075 13.— $1188.52 

7.— 90.1066379 14.— $113.39 

SUBTRACTION OF DECIMALS. 

Lesson 30. Page 30. 

1.— 3.379 8.-24.9975 

2.-6.2236 9.-99.99 

3.-4.784 10.— 33.393 

4.— 0.00594 11.— 1.6483 

5.— 4.79075 12.— 16.85835 

6.— 5.9501 13.— 0.002711 

7.-29.87633 14.— 5.9926 

MULTIPLICATION OF DECIMALS. 

Lesson 31. Page 32. 

1.— 41.82 9.-6.29629 

2.-85 10.— 0.20397 

3.— 121.5 11.— 1500 balls 

4.— 12.15 12.— 13. 3518 inches 

5.— 1.215 13.— $694,500 

6.— 182.8948 14.— $9.02 

7.-28.2744 15.— $2.15 

8.-10,991 . 5 16.— 838-8 inches 

DIVISION OF DECIMALS. 

Lesson 32. Page 34. 

1.— 8.257 15.— 61.386 

2.— 13.367 16.— 0.25 

3.-4.448 17.— 8 

4.-4.857 18.— 6.0024 

5.— 22.105 19.— 0.1016 

6.— 0.378 20.— 0.101419 

7.— 1.443 21.-8.6^ 

8.— 0.00962 22.-24.3^ 

9.-2 . 603 23.-28 . 644 inches 

10.— 0.0015 24.— 17. 2 ins. 22.918 ins. 

11.— 2.004 25.— 3. 75 ins. 



LUDLOW TEXTILE ARITHMETIC 109 



12.— 0.006 




26,— 11. 46 ins. 


13.— 298.853 




27.— 39.37 min. 


14.— 0.02608 




28.— 120 revs. 


EDUCTION OF 


COMMON 


FRACTIONS TO DECIMALS. 




Lesson 33 


. Page 35. 


1.— 0.5 




6.— 0.3 


2.— 0.25 




7.— 0.75 


3.— 0.2 




8,-0.5 


4.^-0.1 




9.— 0.125 


5.— 0.4 




10.— 0.1666 


REDUCTION OF COMMON FRACTIONS TO DECIMALS. 


Lesson 34. 


Page 35-36, 


1.— 0.375 




11.— 1.1875 


2.— 0.4375 




12.— 5.75 


3.— 0.5625 




13.— 11.4062 


4.— 0.625 




14.— 40.4531 


5.— 0.6875 




15.— 18.4687 


6.— 0.15625 




16.— 9.9844 


7.— 0.875 




17.— 8.44 


8.— 0.28125 




18.— 2.3958 


9.— 0.171875 




19.— 0.8298 


10.— 0.390625 




20.— 0.9734 



MISCELLANEOUS EXAMPLES IN DECIMALS. 

Lesson 35. Page 36-37-38. 

1.— 700 pounds 16.— $7250 

2.— . 625 17.— 1 1 cents 5 mills 

3.-2638 . 944 feet 18,— 400 reels 

4.— . 1964 19.— 316 bobbins 

5.— $1 . 38 20.— 840 cuts 

6.— 502 revs. 21.— 38620 . 8 spyndles 

7.-48 lags 22.— 14510 . 4 spyndles 

8.— $2 . 36 23.— 10012 . 8 spyndles 

9.— $2 . 24 24.-5 . 5 pounds 

10.— $1 . 36 25.-335 . 5 H. P. 

11.— 78.25 lbs. 26.-39 H. P. 

-12.— $0.11 27.— 0.87 ounces - 

13.— 243 grains 28.-2 . 75 cuts 

14.— 675 lbs. 29.— 200 R. P. M. 

15.— $4 . 68 30.— 5616 revs. 



110 



LUDLOW TEXTILE ARITHMETIC 
PROPORTION. 



Lesson 36. Page 39. 



1.— 10 to 5 
2.-5 to 7 
3.-3 to 5 
4.-3 to 4 
5.— 100 



Lesson 37. Page 40-41. 



1.— 15 1/2 lbs. 
2. — 131 grains 
3.-32 draft wheel 
4.— 31 draft wheel 
5. — 43 draft wheel 
6.— 2011 R. P. M. 
7.— 140 4/10 bales 



Lesson 38. Page 41. 



1.— 5012 4/5 lbs. 
2. — 6 inch diam. 
3.-24.3 gr. 
4'.— 340.2 gr. 
5.— 121.5 gr. 



6.-243 gr. 
7.— 364.5 gr. 
8.-486 gr. 
9.— 607.5 gr 
10.— 729 gr. 



Example 1. 

1.— 24.3'gr. 

2.-49 gr. 

3.-73 gr. 

4.-96 gr. 

5.— 121 gr. 

6.— 146 gr. 

7.— 170 gr. 

8.— 194 gr. 

9.— 219 gr. 
10.— 243 gr. 
11.— 267 gr. 
12.— 292 gr. 
13.— 316 gr. 
14.— 340 gr. 
15.— 364 gr. 



Lesson 39. Page 41-42. 



16.— 389 gr, 
17.— 413 gr. 
18.— 437 gr. 
19.— 462 gr. 
20.— 486 gr, 
21.— 510 gr. 
22.-534 gr. 
23.— 560 gr. 
24.-583 gr. 
25.— 607 gr 
26.-632 gr, 
27.-656 gr. 
28.— 680 gr. 
29.— 705 gr. 
30.— 729 gr. 



LUDLOW TEXTILE ARITHMETIC 



111 



Exampl 


3 2. 










1.— 3.1416 




7.— 21.9912 


2.-6.2832 




8.— 25.1328 


3.-9.4248 




9.-28.2744 


4.— 12.5664 




10.— 31.416 


5.— 15.708 




11.— 34.5576 


6.— 18.8496 




12.— 37.6992 




Lesson 40. 


Page 42. 


Example 1. 






1 lea weighs 


1166 grains 




2 ' 






583 ' 




- 


3 ' 






389 ' 






4 ' 






292 ' 






5 ' 






233 ' 






6 ' 






194 ' 






7 ' 






174 ' 






8 ' 






146 ' 






9 ' 






130 ' 






10 " 




117 ' 






2.-288 yds. 




6.— 2814 R. P. M 


3.-47 wheel 




7.-8.4 draft 


4.-20,000 bales 


8. — 18.9 inches 


5.— $5641 












FRACTIONS. 



Lesson 41. Page 44. 

1.— 7/3 6.-44/5 

2.-7/2 7.-44/7 

3.— 19/5 8.— 101/10 

4.— 503/100 • 9.— 100/8 

5.-47/8 10.— 100/9 

Lesson 42. Page 44. 
1.— 55/8 • 6.— 331/14 

2.-32/9 7.-989/22 

3.-38/3 8.— 200/3 

4.— 192/11 9.— 175/2 

5.— 119/6 10.— 6279/64 

REDUCTION OF FRACTIONS. 

Lesson 43. Page 45. 
1.— 1 1/4 7.-5 1/10 

2.-2 1/6 8.-6 1/7 

3.-5 2/3 9.-7 2/5 



112 



LUDLOW TEXTILE ARITHMETIC 



4.-2 1/11 


10.— 4 1/3 


5.-3 9/13 


11.— 3 1/5 


6.-3 1/12 


12.— 5 5/8 


Lesson 44. 


Page 45. 


1.— 3 1/12 


10.— 11 46/47 


2.-2 17/20 


11.— 9 


3.-5 


12.— 1 


4.-3 


13.— 43 10/47 


5.-2 1/19 


14.— 10 69/100 


6.-5 11/23 


15.— 17 38/43 


7.— 21 11/12 


16.— 1 35/94 


8.-7 48/53 


17.— 3 


9.-8 10/29 


18.— 4 



REDUCTION OF FRACTIONS TO GIVEN 
DENOMINATOR. 



—99/11 
—60/12 



—144/12 

—60/4 

—100/5 

—48/3 
—34/2 



Lesson 45. Page 45. 

9.-72/3 
10.— 126/6 
11.— 100/4 
12.— 130/5 
13.— 320/8 
14.— 84/2 
15.— 300/6 



REDUCTION OF FRACTIONS TO LOWEST TERMS. 



Lesson 46. Page 46. 



1.— 1/4 
2.-8/9 
3.-6/7 
4.— 1/3 

5.— 1/5 
6.— 1/5 



1.— 3/5 
2.— 21/25 
3.— 1/4 
4.— 211/243 

5.-37/77 



7.- 


-2/5 


8.- 

9- 
10.- 
11.- 
12.- 


-3/4 

-15/16 

-3/8 

-3/4 

-9/20 


'agi 


s 46. 


6.- 

7.- 
8.- 


-1/3 

-6/11 

-1/12 


9.- 
10.- 


-429/1190 
-33/40 



LUDLOW TEXTILE ARITHMETIC 



113 



REDUCTION OF FRACTIONS TO HIGHER TERMS. 





Lesson 


48. 


Page 47. 


1.— 4/8 






7.— 90/100 


2.— 8/12 






8.-9/48 


3.— 18/30 






9.-72/99 


4.-8/28 






10.— 16/36 


5.— 9/60 






11.— 15/51 


6.— 21/45 






12.— 18/39 


m 


Lesson 


49. 


Page 47. 


1.— 84/144 






7.— 440/460 


2.— 78/143 






8.— 570/600 


3.— 105/112 






9.— 78/198 


4.— 112/400 






10.— 85/210 


5.— 81/234 






11.— 77/336 


6.— 40/135 






12.— 240/260 


MULTIPLY A FRACTION BY A WHOLE 




Lesson 


50. 


Page 48. 


1.— 6/7 






7.-6 


2.-2 1/4 






8.-2 1/2 


3.-4 






9.— 1 3/4 


4.-2 1/3 






10.— 12/17 


5.-5 1/2 






11.— 24 


6.-4/7 






12.— 10 




Lesson 


51. 


Page 48. 


1.— 51 






7.-48 2/5 


2.-222 3/4 






8.-364 


3.— 405 






9.— 145 2/3 


4.-224 






10.— 66 


5.— 71 2/5 






11.— 68 


6.— 140 






12.— 372 



MULTIPLY A FRACTION BY A FRACTION. 

Lesson 52. Page 49. 



1.— 1/3 

2.-9/35 
3.-6/55 

4.— 5/14 

5.— 12/77 
6.— 1/6 



7.— 7/20 
8.— 18/35 
9.— 8/11 
10.— 1/30 
11.— 3/4 
12.— 1/2 



114 



LUDLOW TEXTILE ARITHMETIC 



MULTIPLICATION OF FRACTIONS. 

Lesson 53. Page 49. 



1.— 2/9 




7.— 1/28 


2.— 1 1/3 




8.— 1/3 


3.— 63/380 




9.— 7/15 


4.-377/945 




10.— 2/9 


5.-4/45 




11.— 1/8 


6.— 17/282 




12.— 3/34 




RECIPROCALS. 




Lesson 


54. Page 49. ■ 


1.— 1/3 




7.-9/7 


2.— 1/6 




8.— 11/5 


3.— 1/7 




9.-23/6 


4.— 1/13 




10.— 7/23 


5.— 1/17 




11.— 13/29 


6.— 1/22 




12.— 6/41 


DIVISION OF FRACTIONS BY WHOLE NUMBEE 




Lesson 


55. Page 50. 


1.— 1/9 




7.— 2/169 


2.— 1/15 




8.-2/45 


3.— 1/10 




9.— 1/30 


4.— 1/14 




10.— 1/27 


5.— 1/12 




11.— 1/18 


6.— 1/33 




12.— 1/64 


DIVISION OF FRACTIONS BY FRACTIONS. 




Lesson 


56. Page 50. 


1,-1 1/5 




8.— 1 


2.— 1 17/108 




9.— 1 3/4 


3.-2 2/7 




10.— 36/49 


4.-9/25 




11.— 507/1000 


5.— 1 79/210 




12.— 2 2/47 


6.— 1 19/27 




13.— 2 77/150 


7.— 33/70 




14.— 1 


FRACTIONAL PART 


OF ANOTHER FRACTION, 




Lesson 


57. Page 51. 


1.— 1/5 




7.— 1/4 


2.— 1/6 




8.— 1/6 


3.— 1/5 




9.— 1/8 


4.— 1/6 




10.— 3/8 


5.— 1/3 




11.— 1/16 


6.— 2/11 




12.— 1/40 



LUDLOW TEXTILE ARITHMETIC 115 

SIMILAR FRACTIONS. 





Lesson 


58. 


Page 52. 




1.- 

2.- 

3.- 
4.- 
5.- 


-3/6, 2/6 
-3/12, 4/12 
-8/40, 5/40 
-4/6, 5/6 
-7/8, 6/8 






6.— 5/16, 10/16 

7.— 18/63, 28/63 
8.— 9/15, 7/15 
9.— 9/30, 20/30 
10.— 40/48, 42/48 






Lesson 


59. 


Page 52. 




1.- 
2.- 
3.- 
4.- 
5.- 


-2/16, 12/16, 
-42/54, 45/54, 
-40/50, 9/50, 
-54/60, 4/60, 
-10/30, 15/30, 


5/16 

28/54 
46/50 
35/60 

12/30 


6.— 14/46, 23/46, 
7.— 28/70, 45/70, 
8.— 28/60, 20/60, 
9.— 18/48, 32/48, 
10.— 25/90, 72/90, 


9/46 

30/70 

25/60 

9/48 
70/90 




ADDITION 


OF 


FRACTIONS. 






Le 


:sson 


60. 


Page 53. 




1.- 
2.- 
3.- 
4.- 
5.- 


-3/8 

-13/16 

-11/16 

-7/8 

-9/16 






6.— 15/16 
7.— 1 5/8 
8.-27/64 
9.— 15/64 
10.— 1 1/64 






Lesson 


61. 


Page 53. 




1.- 
2.- 
3- 

4.- 
5.- 


-2 9/16 
-37/50 
-1 1/5 
-2 7/24 
-71 11/16 






6.— 101 7/48 
7.-34 37/45 
8.-2 59/84 
9.— 51 11/16 
10.— 21 5/12 






SUBTRACTION OF FRACTIONS. 






Lesson 


62. 


Page 54. 




1.- 
2.- 

3.- 
4.- 
5.- 


-1/4 
-2/5 
-3/8 
-3/16 

-7/24 






6.— 5/16 
7.-4 1/4 
8.— 21 7/16 
9.-26 3/8 
10.— 35 53/64 






Lesson 


63. 


Page 54. 




1.- 
2.- 

3.- 
4.- 
5.- 


-23 3/8 
-22 1/4 
-20 9/16 
-15 1/8 
-33 13/16 






6.-9 57/64 
7.-27/64 
8.-2 7/100 
9.-3 31/40 
10.— 37/60 





116 



LUDLOW TEXTILE ARITHMETIC 



CHANGE A DECIMAL TO A COMMON FRACTION. 

Lesson 64. Page 55. 



1.- 


-1/5 




11.— 3 24/25 


2.- 


-3/5 




12.— 6 333/500 


3.- 


-1/2 




13.— 7 1/8 


4.- 


-1/4 




14.— 4 1/16 


5.- 


-3/4 




15.— 12 12/125 


6.- 


-61/200 




16.— 1/2000 


7.- 


-3/50 




17.— 14 9/1000 


8.- 


-3/500 




18.— 7 3/8 


9.- 


-649/2000 




19.— 1/4000 


10.- 


-501/1250 




20.— 10 2119/5000 




MISCELLANEOUS EXAMPLES. 




Lesson 


65. 


Page 55-56. 


1.- 


-5/12 




7. — 5/8" cents 


2.- 


-$10 




8.-85 1/4 bales 


3.- 


-$23 




9.-27 1/2 doffs 


4.- 


—75 cents 




10.— 59 3/8 cents 


5.- 


-92 1/2 cents 




11.— 46 balls 


6.- 


-21 bales 




12.— 358 balls 




Lesson 


66. 


Page 56-57. 


1.- 


—347 1/3 pounds 




11.— 0.1964 inches 


2.- 


-$918.75 




12.— $0,761 


3.- 


-$18,125 




13.— $0.25 


4.- 


-$37 . 33 




14.— $0,814 


5.- 


—1162 pounds 




15.— $5.53 


6.- 


-0.60416 ton 




16.— $54.58 


7.- 


-$8.21 




17.— 106 balls 


8.- 


-$112 




18.— 24 7/8 cents 


9.- 


-$18.24 




19.— $6.00 


10.- 


— 6 yards 








WEIGHTS 


! AND MEASURES. 




YARN EXAMPLES. 




Lesson 


67. 


Page 60-61. 


1.- 


—486 gr. 




10.— 280 gr. 


2.- 


—4 lea 




11.— 64.3 gr. 


3.- 


—40 threads 




12.— 450 yds. 


4.- 


—292 gr. 




13. — 6 ozs. 


5.- 


—10 lbs. 




14.— 6000 spyndles 


6.- 


—3 lea 16 lbs. 




15.— 5119 cuts 



LUDLOW TEXTILE ARITHMETIC 



117 



7.-583 gr. 
8.— 30 lbs. 
9.— 25's lea 



19.— 

1 lb. = 

2 1bs.= 

3 1bs.= 
41bs.= 

5 lbs. = 

6 lbs. 
Tibs. 

8 lbs. 

9 lbs. 
10 lbs. 



49 gr. 

98 gr, 

146 gr. 

194 gr 

243 gr 

= 292 gr 

= 340 gr 

= 389 gr, 

= 437 gr, 

= 486 gr, 



16.— 3840 cuts 
17.— 8000 spyndles 
18. — 36 spyndles 
20.— 

l's = 1166 gr. 

2's = 583gr. 

3's = 389 gr. 

4's = 292gr. 

5's = 233gr. 

6's = 194gr. 

7's = 174 gr. 

8's = 146gr. 

9's = 130 gr. 
. 10's = 117 gr. 



LONG MEASURE. 

Lesson 68. Page 63. 



1. — 231 inches 

2.— 179 inches 

3. — 12 yards 

4.— 11 yds., ft., 1 in. 

5.— 3 mi., 39 rds., 2 yds. 

6. — 6 mi., 126 rds., 1 yd. 

7.— 270 inches 



2.5 ft. 

2 ft. 



WEIGHTS AND MEASURES. 

Lesson 69. Page 64. 



1.— 12 yds., 1 ft., 10 ins. 
2.-49 yds., 1 ft., 3 ins. 
3.-23 mi., 1623 yds. 
4.— 20 mi., 1015 yds., 2 ft. 
5.— 11 yds., 1 ft., 10 ins. 
6.-8 yds., 2 ft., 8 ins. 
7.-2 mi., 1366 yds., 2 ins. 
8.— 6 mi., 1756 yds., 1 in. 



LONG MEASURE. 

Lesson 70. Page 65. 



1.— 186 yds. 

2.— 319 yds., 1 ft., 4 ins. 
3.— 18 mi., 804 yds., 2 ft. 
4.— 61 mi., 229 yds. 
5.-22 yds., 3 7/8 ins. 
6.— 19 yds., 2 ft., 7 3/4 ins. 



118 LUDLOW TEXTILE ARITHMETIC 

Lesson 71. Page 65. 
1.— 2352 feet 5.-98,181 . 8 miles 

2.— 1450 2/3 yards 6.— 11 spyndles 

3.— 6035 rails 7.— 108 lengths 

4.-255,200 yards 8.— 500 yards 

SQUARE MEASURE. 

Lesson 72. Page 66. 
1.-79,200 sq. ft. 6.-36 sq. yds. 

2.-17,424 sq. ft. 7.— 2404 sq. yds. 

3.-6453.33 sq. yds. 8.— 12 sq. yds. 

4.-196,020 sq. ft. 9.— 108.8 sq. yds. 

5. — 66 sq. yds. 

CUBIC MEASURE. 

Lesson 73. Page 66-67. 
1.— 96 cu. ft. 6.— 210.54 63/108 cu. ft. 

2.— 126 cu. ft. 7.-185,856 cu. ft. 

3.— 1800 cu. ft. 8.-^6666 + bales 

4.-278,784 cu. ft. 9.-72 yds. 

5.— 802. 12 cu. ft. 10.-41,472 cu. ins. 

LIQUID MEASURE. 

Lesson 74. Page 67-68. 
1.— 45 gals., 2 qts., 1 pt. 6.— 3715 gals., 1 qt. 



2.-67 gals., 1 qt., 1 pt 

3. — 5 gals., 3 qts., 1 pt. 8 

4.-8 gals., 2 qts., 1 pt. 9 

5.-46,080 pts. 10 



69 gals. 
—107 gals., 2 qts., 1 pt. 
— 29 gals., qts., 1 pt. 
—27 gals., qts., 1 pt. 



AVOIRDUPOIS WEIGHT. 

Lesson 75. Page 68. 
1.— 1312.5 gr. 6.-50,000 tons 

2.-75,900 lbs., oz., 3 gr. 7.— 1280 tons 
3.-248 lbs., 6684 gr. 8.— 128 ozs. 

4.-98,000 gr. 9.— 4000 ft. 

5.— 17 tons, 1918 lbs., 4 ozs. 10.— $0,075 

MISCELLANEOUS. 

Lesson 76. Page 70-71. 

1.— 6104 . 67 bales 9.-5985 ft. 

2.— 11 . 32 systems 10.— 213 . 55 gals. 

3.-253,440 feet 11.— 2184 lbs. 

4. — 8712 storehouses 12. — 475 bobbins 



LUDLOW TEXTILE ARITHMETIC 119 

5.-14,840 . 32 gals. 13.— 324 cwt. 

6.-41,184 feet 14.— 261 lbs., 4 ozs. 

7.— The first 3816 cu. ins. 15.— $2,880,000 
8.-14,784 lbs. 

ALLIGATION. 

Lesson 77. Page 72. 
1.— 7.1jzf 
2.-11^ 
3.— 11 5/8^ 
4-— 12jzT 
5.— 10 2/3^ 

Lesson 78. Page 73. 
1. — 2 flax, 3 hemp 
2. — Equal parts 
3. — 2 flax, 1 hemp 
4. — Infinite number of answers 
5. — Infinite number of answers 

PERCENTAGE. 
A COMMON FRACTION AS A RATE PER CENT. 

Lesson 79. Page 75. 
1.-25% 7.-31.25% 

2.-33.33% 8.-41.65% 

3.-50% 9.-42.86% 

4.-16.66% 10.-44.44% 

5.-12.5% 11.-22.5% 

6.-10% 12.-143.75% 

A DECIMAL AS A RATE PER CENT. 

Lesson 80. Page 75. 

1.-46% 7.-648% 

2.-38% 8.— .05% 

3.-7% 9.-1004% 

4.-90% 10.-0.625% • 

5.-26% 11.-12.5% 

6.-779% 12.-3.75% 

FIND PERCENTAGE HAVING BASE AND RATE 
GIVEN. 

Lesson 81. Page 76. 

1.— 15 lbs. 5.— 1250 lbs. 

2.-32 lbs. 6.— 2200 lbs. 

3.-62 . 5 lbs. 7.— 2800 lbs. 
4.— 200 lbs. 



120 LUDLOW TEXTILE ARITHMETIC 

PERCENTAGE. 

Lesson 82. Page 76. 

1.— 2160 spindles 

2.— 130 lbs. 

3.— 640 lbs. 

4.-60,000 lbs. 

5.-96 looms, 4632 yards 

FIND RATE HAVING BASE AND PERCENTAGE 

Lesson 83. Page 77. 

1.-5.43% 7.— 1897 3. 17% increase 
2.-4.12% 1898 

3.-2% 1899 11.38% increase 

4._5.92% 1900 23 . 76% increase 

5—37% 1901 8.70% increase 

6. — 23% 1902 7.39% increase 

1903 1 . 55% decrease 

1904 9 . 57% decrease 

1905 23.00% decrease 

1906 5 . 02% increase 

1907 . 33% increase 

1908 19.50% increase 

TO FIND BASE. 

Lesson 84. Page 78. 

1.— $225 5.— 8800 

2.— $1250 6.— 300 

3.— 400 7.— 600 
4.-625 

Lesson 85. Page 78. 
1.— 1435 lbs. 4.— $6 

2. — 250 lbs. 5. — 705 children 

3.-12,000 lbs. 6.— 20317. 5 lbs.' 

PERCENTAGE. 

Lesson 86. Page 78-79. 

1.— 60 bales 
2. — 250% increase 
3. — 59.12% increase 
4. — 112.62% increase 
5. — 51 . 5% increase 



6.- 


-6.577 


7.- 


-0 . 186 


8.- 


-0.871 or 27/31 


9.- 


-0 . 689 or 15/67 


0.- 


-0.831 



LUDLOW TEXTILE ARITHMETIC 121 

SQUARE ROOT. 

Lesson 87. Page 82. 

1.— 351 
2.-89 . 7 
3.— 7008 
4.— 10.5 
5.— 9812 

GENERAL MILL WORK. 

- DIAMETERS AND CIRCUMFERENCES. 

Lesson 88. Page 83. 
1.— 7 . 0686 inches 4.— 15 . 708 feet 

2.-7 inches 5.-28 . 68 inches 

3.— 88 inches ; 6.— 2.353 inches 

LENGTH OF BELL. 

Lesson 89. Page 84. 
1. — 175 yards 
2.— 140 yards 
3.— 155 1/2 yards 
4. — 176 yards 

TRACING SPEED. 

Lesson 90. Page 85. 
1.— 437.5 R. P. M. 
2.-233.3 R. P. M. 
3.— 131.25 R. P. M. 
4.— 70 R. P. M. 

SURFACE SPEED. 

Lesson 91. Page 86. 
1. — 473 . 2 feet per minute 
2. — 30 . 628 feet per minute 
3. — 104.3 feet per minute 
4. — 100 revolutions per minute 
5. — 5024 feet per minute 

TWISTING. 

Lesson 92. Page 95. 
1. — 4.6 turns per inch 
2. — 4 . 9 turns per inch 
3. — 3 . 7 turns per inch 



122 



LUDLOW TEXTILE ARITHMETIC 



.— l's = 2.0 turns 


per inch 


2's = 2.8 


< i 






3's = 3.4 


i i 






4's = 4.0 


<< 






5's = 4.5 


ii 






6's=4.9 


" 






7's = 5.3 


" 






8's = 5.8 


(< 






9's = 6.0 


" 






10's = 6.3 


ii 


ii i 




. — 1 pound 


yarn - 


13.8 turns per inc 


2 pounds 


" 


9.9 " " " 


3 


ii 


8.0 " " " 


4 


ii 


6.9 " " " 


5 


1 1 


6.1 " " " 


6 


ii 


5.6 





POUNDS PER SPYNDLE. 

Lesson 93. Page 100. 



1. — 45 pounds per spyndle 
2. — 108 pounds per spyndle 
3. — 63 pounds per spyndle 
4. — 144 pounds per spyndle 
5, — 90 pounds per spyndle 





DRAFT ON SPINNING FRAME. 






Lesson 94. Page 100. 


1.- 


-10.5 draft 5.-2.7 


2.- 


-12.6 


6.— 5.0 


3.- 


-5.4 


7.-4.5 


4.- 


-5.8 


SPINNING DRAFT. 

Lesson 95. Page 101. 


1.- 


—9 draft 




2.- 


-6.9 




3.- 


-6.4 




4.- 


-6 


DRAFT. 

Lesson 96. Page 101. 


1.- 


-5.1 draft 


4.-8.3 draft 


2.- 


-6.8 


5.-3.8 


3.- 


-6.8 


6.-2.3 



OCT 131909 



